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Geometry with coordinates
A STORY
Two discs on the tiles
Kabir kneels on the tiled terrace beside a pale disc. Ananya stands a few tiles away and sets a
brass disc down on the floor.
“Count the tiles to each disc, from that corner,” Ananya says.
Kabir counts to the pale disc. “$3$ across, $2$ up.” He counts to the brass disc. “$7$ across,
$6$ up.”
“Two numbers each, and you know exactly where both discs sit,” Ananya says.
“But how far apart are they?” Kabir asks. “I do not want to lay a tape across the whole floor.”
“You do not need to,” Ananya says. “The tile counts alone give you the distance, without ever
touching a tape to the floor.”
“Just from four numbers?” Kabir asks.
“Just from four numbers,” Ananya says.
You will learn the rule that turns those four numbers into one distance. It works for any two
points on a floor like this.
Four counts fix two points, and the distance formula turns those four counts into one length.
Every point on a flat map can be pinned down by two numbers. Once it has those two numbers, we can find the distance between two points without a ruler.
Picture a district grid. Streets run east to west and north to south. Each street is numbered out from a fixed corner. Two towns sit at two of those grid points, joined by one straight road. Walk from one town to the other and you cross $3$ streets going east and $4$ going north, and the road itself is $5$ long. One rule does that for any two towns, and the next section builds it.
A point can also sit ON that road, instead of being measured across it. Cut the road so that one piece is twice as long as the other, and ask where the cut falls. The same handful of numbers answers that too, through a second rule, a few pages further on.
Two jobs, one coordinate picture. Every section from here on puts one of them to work.
Your turn: two towns sit at $(2, 5)$ and $(9, 5)$. How far apart are they? (Answer: $7$ units, since they share a $y$-coordinate, and the gap is $9 - 2$.)
A ruler laid along the plotted segment reads 5 units, and the two coordinate pairs give the same 5 with nothing drawn at all.
A SCHOLAR INDIA REMEMBERS
Brahmagupta worked in a school of astronomy with a careful habit. A calculation said where something would be in the sky. The astronomers checked it against the sky. When the two did not match, they corrected the numbers in the calculation. They did not argue with what they saw. You can do the same here. Work out a distance with the formula, then look at the graph. If the two disagree, fix the working.
Brahmagupta · the brahminy kite
Check yourself
With points written as coordinate pairs, this chapter’s opening claim hands you two results directly, once the points are known. What are they?
the slope of the line through two points
a length, and a dividing point
the area enclosed by three or four points
the equation of the line joining two points
Check your answer
the slope of the line through two points — The frame never mentions a slope — this chapter builds only a length and a dividing point.
✓ a length, and a dividing point — (B) The frame states exactly two results: a length between two points, and a new point from a ratio.
the area enclosed by three or four points — Area-of-a-triangle by coordinates is not part of this chapter’s frame — only distance and division survive.
the equation of the line joining two points — This chapter never writes a line’s equation — it only measures a length and locates a dividing point.
Both formulas in this chapter’s frame — the length between two points and the dividing point for a ratio — are built from the same two ingredients. What are those ingredients?
just the two points’ $x$- and $y$-coordinates
the two points’ coordinates plus the slope of the segment joining them
the two points’ coordinates plus the angle the segment makes with the $x$-axis
a diagram drawn to scale, read off with a ruler
Check your answer
✓ just the two points’ $x$- and $y$-coordinates — (A) Both formulas read straight off the coordinates of the two points; nothing else is needed.
the two points’ coordinates plus the slope of the segment joining them — Neither formula in this chapter ever computes a slope — coordinates alone are enough.
the two points’ coordinates plus the angle the segment makes with the $x$-axis — No angle appears in either formula — both work from coordinates alone.
a diagram drawn to scale, read off with a ruler — Neither formula needs a scale drawing — both are computed directly from the coordinates.
This chapter’s two formulas feed a bigger task: deciding a shape’s type from its vertices. Which task below needs only the distance formula, never the section formula?
finding the point that divides a diagonal in the ratio $2 : 3$
classifying a shape by its side lengths
finding the midpoint of a diagonal
finding the ratio in which a known point divides a segment
Check your answer
finding the point that divides a diagonal in the ratio $2 : 3$ — Dividing a segment in a given ratio is the section formula’s job, not the distance formula’s.
✓ classifying a shape by its side lengths — (B) Classifying a shape by its sides needs only lengths, so only the distance formula is used.
finding the midpoint of a diagonal — A midpoint comes from the section formula (with $m = n$), not from the distance formula alone.
finding the ratio in which a known point divides a segment — Finding a ratio from a known point still runs the section formula, only in reverse.
You have plotted points before. Now measure between them. Try each check below. It takes a minute.
A point’s two distances (Class 9, Ganita Manjari, page 14). Here: every point’s $x$ and $y$ are read off as these two distances. Check: Point $(-4, 7)$ sits $4$ units from the $y$-axis and $7$ from the $x$-axis.
The Pythagoras theorem (Class 8, Ganita Prakash Part 2, page 87). Here: the distance formula applies this same theorem on the grid. Check: Legs $24$ and $32$ give hypotenuse $\sqrt{24^2 + 32^2} = 40$.
Dividing a whole in a ratio (Class 8, Ganita Prakash Part 2, page 58). Here: the section formula divides a segment in a given ratio, just like this. Check: $12$ divided in ratio $2:1$ gives $8$ and $4$, since $8:4$ is $2:1$.
A ratio in simplest form (Class 8, Ganita Prakash Part 1, page 162). Here: a found ratio here is confirmed by reducing it to simplest form. Check: $4:2$ and $10:5$ both reduce to $2:1$, so they are in proportion.
Equal side lengths (Class 8, Ganita Prakash Part 1, page 99). Here: a figure here is named from its equal sides, just like this one. Check: A quadrilateral with all four sides $5$ has perimeter $4 \cdot 5 = 20$.
Squaring a sum (Class 9, Ganita Manjari, page 70). Here: each distance, squared to drop its root, expands like this. Check: $(3 + 4)^2 = 3^2 + 2 \cdot 3 \cdot 4 + 4^2 = 49$.
If any of these felt new, read the page named before going on.
A square on a sum splits into two squares and two rectangles, and that expansion is what every squared distance in this chapter unpacks into.
A point in a plane is pinned down by an ordered pair of coordinates, $(x, y)$.
*The $x$-coordinate is the point’s distance from the $y$-axis. The $y$-coordinate is its distance from the $x$-axis.* Get this pairing backwards, and every formula built on it points at the wrong spot.
Try it yourself before reading on. What are the coordinates of a point that sits $4$ units from the $y$-axis, and none from the $x$-axis? Such a point sits on the $x$-axis itself, at $(4, 0)$. Now swap the reasoning. $(0, 4)$ is a different point, sitting on the $y$-axis instead.
The naming looks backwards until we have used it a few times. The axis a coordinate is measured FROM is not the axis that shares its name.
Your turn: a point sits $3$ units from the $x$-axis and $7$ units from the $y$-axis, in the first quadrant. What are its coordinates? (Answer: $(7, 3)$, since $x$ is measured from the $y$-axis and $y$ from the $x$-axis.)
The point at 5 and 3 has dashed lines showing 5 as its distance from the y-axis, and 3 from the x-axis.
A SCHOLAR INDIA REMEMBERS
Where is $(-3, 2)$? Is $-3$ a real place to stand? Yes, it is. Start at the origin. Move $3$ units to the left, because the number is negative. Then move $2$ units up. A negative coordinate is still a place. It is on the other side of the origin.
Brahmagupta · the brahminy kite
Check yourself
For a point $(x, y)$, the $x$-coordinate is the point’s distance from
the $y$-axis
the $x$-axis
the origin
Check your answer
✓ the $y$-axis — (A) The $x$-coordinate measures how far the point sits from the $y$-axis.
the $x$-axis — The $x$-coordinate is measured from the $y$-axis, not the $x$-axis — that distance is the $y$-coordinate’s job.
the origin — Distance from the origin uses both coordinates together — the $x$-coordinate alone measures distance from the $y$-axis only.
A student says: “for the point $(3, 5)$, the $3$ tells me how high up it is, and the $5$ tells me how far right.” What is wrong with this?
nothing is wrong — the student has it right
both numbers actually measure the same distance, from the origin
the two are swapped
the point should be written as $(5, 3)$ instead
Check your answer
nothing is wrong — the student has it right — The claim has the two coordinates swapped: $x$ measures across, $y$ measures up, not the reverse.
both numbers actually measure the same distance, from the origin — $x$ and $y$ measure two separate distances, from two separate axes — neither alone is the distance from the origin.
✓ the two are swapped — (C) The coordinates are swapped in the student’s claim: $x$ measures across, $y$ measures up.
the point should be written as $(5, 3)$ instead — The point $(3, 5)$ is written correctly — the fix is in what each coordinate is said to measure, not in the point itself.
Why is the $x$-coordinate measured from the $y$-axis, and not from the $x$-axis?
because the $y$-axis is always drawn on the left of a graph
it tracks distance from the $y$-axis
because $x$ comes before $y$ in the alphabet
because the $x$-axis only measures whole numbers
Check your answer
because the $y$-axis is always drawn on the left of a graph — How a graph happens to be drawn is not why a coordinate measures a distance — the axis itself defines it.
✓ it tracks distance from the $y$-axis — (B) The $x$-coordinate tracks left-right movement, which is distance from the $y$-axis.
because $x$ comes before $y$ in the alphabet — Alphabetical order has no bearing on which axis a coordinate is measured from.
because the $x$-axis only measures whole numbers — Neither axis is restricted to whole numbers — this is not why $x$ is measured from the $y$-axis.
A point $P$ divides the segment joining $A$ and $B$ internally when $P$ sits between the two points, on the straight road connecting them.
*The ratio $m : n$ compares $P$’s distance from $A$ to its distance from $B$, in that order.* Picture a ratio of $1 : 1$. It places $P$ exactly halfway. Now picture $1 : 4$ instead. $P$ sits close to $A$, since its share of the distance is the smaller number.
Check this on any ratio. $P$ must sit between $A$ and $B$ for the division to be internal. A point outside the segment divides it externally instead.
Your turn: $P$ divides the segment joining $A$ and $B$ in the ratio $1 : 3$. Which point does $P$ sit closer to? (Answer: $A$, since the ratio names $P$’s distance from $A$ first, and $1$ is the smaller share.)
Two tick marks on one piece of the segment, and three on the other, show the ratio 2 to 3, with no coordinates yet.
Check yourself
Point $P$ divides segment $A B$ internally in the ratio $3 : 2$. This means
$P$ sits exactly $3/2$ of the way along $A B$, measured from $A$
$P$’s distance from $A$ compares to its distance from $B$ as $3$ is to $2$
segment $A P$ is always exactly $3$ units and $P B$ is always exactly $2$ units
$P$ lies exactly at the midpoint of $A B$
Check your answer
$P$ sits exactly $3/2$ of the way along $A B$, measured from $A$ — $P$ sits $3/(3+2) = 3/5$ of the way from $A$ to $B$ — not $3/2$, which is larger than the whole segment.
✓ $P$’s distance from $A$ compares to its distance from $B$ as $3$ is to $2$ — (B) The ratio compares $P$’s distance from $A$ to its distance from $B$, as $3$ to $2$.
segment $A P$ is always exactly $3$ units and $P B$ is always exactly $2$ units — The ratio $3 : 2$ fixes a proportion between $A P$ and $P B$, not a fixed length of $3$ and $2$ units.
$P$ lies exactly at the midpoint of $A B$ — A ratio of $3 : 2$ is not $1 : 1$, so $P$ is not the midpoint of $A B$.
A student claims: “in the ratio $1 : 3$, point $P$ must be closer to $B$ than to $A$, since $3$ is the bigger number.” What is wrong with this?
nothing is wrong — $P$ is indeed closer to $B$
the ratio needs both numbers to be equal before you can compare distances at all
$P$ is actually closer to $A$
$P$ must lie outside the segment $A B$ entirely
Check your answer
nothing is wrong — $P$ is indeed closer to $B$ — $A P : P B = 1 : 3$ makes $A P$ the smaller share, which places $P$ closer to $A$, not $B$.
the ratio needs both numbers to be equal before you can compare distances at all — A ratio compares two distances whether or not its numbers are equal — $1 : 3$ compares them just as validly as $1 : 1$.
✓ $P$ is actually closer to $A$ — (C) $A P : P B = 1 : 3$ makes $A P$ the smaller share, so $P$ sits closer to $A$, not $B$.
$P$ must lie outside the segment $A B$ entirely — “Divides internally” places $P$ between $A$ and $B$ by definition — this ratio gives no reason to put it outside.
Why must a ratio $m : n$ name which point is $A$ and which is $B$, rather than just naming two numbers?
it does not actually matter — the numbers $m$ and $n$ mean the same thing either way
naming $A$ and $B$ is only a labelling convenience, with no effect on the resulting point
swapping $A$ and $B$ swaps what $m$ and $n$ measure
a ratio only needs the two numbers; which point is which is decided by the diagram, not the labels
Check your answer
it does not actually matter — the numbers $m$ and $n$ mean the same thing either way — Swapping which point is $A$ and which is $B$ changes which distance $m$ and $n$ each measure — it does matter.
naming $A$ and $B$ is only a labelling convenience, with no effect on the resulting point — The labels are not just convenience — swapping them moves the resulting point, unless $m = n$.
✓ swapping $A$ and $B$ swaps what $m$ and $n$ measure — (C) Swapping $A$ and $B$ changes which distance each of $m$ and $n$ measures.
a ratio only needs the two numbers; which point is which is decided by the diagram, not the labels — A ratio is defined by which point each number is measured from — a diagram alone cannot supply that.
Two towns on a district grid sit at $(x_1, y_1)$ and $(x_2, y_2)$. What is the straight-line distance between them?
We draw a horizontal line through one town, and a vertical line through the other. The two lines meet at a third point. The three points form a right triangle. Its two legs measure $|x_2 - x_1|$ and $|y_2 - y_1|$, one horizontal and one vertical. The road joining the two towns is the hypotenuse of that triangle.
The Pythagoras theorem gives us that hypotenuse directly. *The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, no matter where the two towns sit.*
What is the distance between $(0, 0)$ and $(3, 4)$? Substitute directly, and you get $\sqrt{3^2 + 4^2} = 5$.
Subtracting inside either bracket in the opposite order changes nothing. $(x_1 - x_2)^2$ and $(x_2 - x_1)^2$ are equal, because squaring removes a negative sign either way.
One more thing the formula settles on its own. Three points are COLLINEAR when they sit on one straight line. Take the three distances between them, two points at a time. If the two shorter ones add up to the longest, there is no bend at the middle point and the three are collinear. If they add up to more than the longest, the middle point sits off the line. Take $(0, 0)$, $(3, 4)$ and $(6, 8)$: the three distances are $5$, $5$ and $10$, and $5 + 5 = 10$, so those three are collinear.
Your turn: find the distance between $(2, 2)$ and $(8, 10)$. (Answer: $10$, since $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$.)
A right triangle on the grid has legs 6 and 8 at a right angle, and a diagonal of 10 as the distance.
Check yourself
The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is
$(x_2 - x_1)^2 + (y_2 - y_1)^2$
$|x_2 - x_1| + |y_2 - y_1|$
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2}$
Check your answer
$(x_2 - x_1)^2 + (y_2 - y_1)^2$ — This leaves out the final square root — it gives the squared distance, not the distance itself.
$|x_2 - x_1| + |y_2 - y_1|$ — This adds the two legs directly (a Manhattan-style distance) instead of applying Pythagoras to them.
✓ $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ — (C) The distance formula squares each coordinate difference, adds them, and takes the square root.
$\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2}$ — The formula subtracts each pair of coordinates before squaring — adding them measures something else entirely.
A student finds the distance between $(2, 3)$ and $(5, 7)$ by writing $\sqrt{3 + 4} = \sqrt{7}$. What went wrong?
nothing — $\sqrt{7}$ is the correct distance
the square root should not have been taken at all
the two points were subtracted in the wrong order
the differences were not squared first
Check your answer
nothing — $\sqrt{7}$ is the correct distance — The differences must be squared first: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$, not $\sqrt{7}$.
the square root should not have been taken at all — The square root is the correct final step — the missing step is squaring the differences before adding them.
the two points were subtracted in the wrong order — Subtraction order makes no difference once the results are squared — the missing step is the squaring itself.
✓ the differences were not squared first — (D) The differences must be squared before adding — $\sqrt{3^2 + 4^2} = 5$, not $\sqrt{3 + 4}$.
Why does the distance formula work?
it averages the two points’ coordinates to estimate how far apart they are
it measures how steeply the segment joining them rises
it counts how many grid squares lie between the two points
they form a right triangle; Pythagoras gives its hypotenuse
Check your answer
it averages the two points’ coordinates to estimate how far apart they are — Averaging the coordinates locates the midpoint between the points — it says nothing about how far apart they are.
it measures how steeply the segment joining them rises — How steeply a segment rises is its slope, a separate idea from how long the segment is.
it counts how many grid squares lie between the two points — The formula computes an exact length algebraically — it does not count grid squares on a drawn graph.
✓ they form a right triangle; Pythagoras gives its hypotenuse — (D) The two points and a right-angle corner form a right triangle whose hypotenuse Pythagoras gives.
The distance formula works whether you compute $x_2 - x_1$ or $x_1 - x_2$. Why does the order not matter?
because distance is always measured moving left to right on a graph
squaring a difference removes its sign, so both orders give the same squared value
because the two points are always interchangeable in every formula in this chapter
because the square root at the end cancels out any sign from the subtraction
Check your answer
because distance is always measured moving left to right on a graph — Distance has no built-in direction to read in — the reason the order is free is that squaring erases the sign.
✓ squaring a difference removes its sign, so both orders give the same squared value — (B) Squaring removes the sign of the difference, so either subtraction order gives the same result.
because the two points are always interchangeable in every formula in this chapter — The section formula’s two points are NOT interchangeable — this order-freedom is special to squaring in the distance formula.
because the square root at the end cancels out any sign from the subtraction — It is the squaring step, not the final square root, that removes the sign — the root only undoes the earlier squaring.
$A(1, 1)$, $C(4, 5)$ and $D(1, 7)$. Which of $C$ or $D$ is closer to $A$?
$C$ is closer
$D$ is closer, since $D$ shares an $x$-coordinate with $A$
they are equally far from $A$
it cannot be decided without a scale drawing
Check your answer
✓ $C$ is closer — (A) $A C = 5$ and $A D = 6$, so $C$ sits closer to $A$.
$D$ is closer, since $D$ shares an $x$-coordinate with $A$ — Sharing an $x$-coordinate with $A$ does not decide closeness — the actual distances, $5$ and $6$, do.
they are equally far from $A$ — The distances are $5$ and $6$ once computed — not equal, and $C$ is the nearer one.
it cannot be decided without a scale drawing — The distance formula settles this exactly — $A C = 5$ and $A D = 6$ — with no drawing required.
The origin $O(0, 0)$ is a point like any other. The distance formula still applies to it directly.
Set $x_1 = y_1 = 0$ in the general formula yourself. One whole term drops out of each bracket. *The distance of a point $(x, y)$ from the origin comes out to $\sqrt{x^2 + y^2}$.* This is the same formula, with one point fixed at the origin.
Check it on $(6, 8)$. You get $\sqrt{6^2 + 8^2} = 10$ units from the origin, found the same way as the distance between any other two points.
Your turn: find the distance of $(7, 24)$ from the origin. (Answer: $25$, since $\sqrt{7^2 + 24^2} = \sqrt{625} = 25$.)
Check yourself
The distance of a point $(x, y)$ from the origin $O(0, 0)$ is
$x^2 + y^2$, leaving the square root off
$\sqrt{x - y}$, subtracting instead of squaring and adding
$\sqrt{x^2 + y^2}$
Check your answer
$x^2 + y^2$, leaving the square root off — This leaves out the final square root — it gives the squared distance from the origin, not the distance itself.
$\sqrt{x - y}$, subtracting instead of squaring and adding — The formula squares each coordinate and adds them — it does not subtract one coordinate from the other.
✓ $\sqrt{x^2 + y^2}$ — (C) This is the distance formula with $x_1 = y_1 = 0$ substituted in.
Why is the origin-distance formula just the distance formula with $x_1 = y_1 = 0$?
the origin is just an ordinary point, $(0, 0)$
because the origin is a special case that needs its own separate rule
because distance from the origin is always smaller than distance between two other points
because the origin lies on both axes at once, so it needs both formulas combined
Check your answer
✓ the origin is just an ordinary point, $(0, 0)$ — (A) The origin is an ordinary point at $(0, 0)$, so the same distance formula applies to it.
because the origin is a special case that needs its own separate rule — No separate rule is needed — substituting $(0, 0)$ into the same distance formula is enough.
because distance from the origin is always smaller than distance between two other points — There is no rule that origin-distances are always smaller — it depends entirely on the specific points.
because the origin lies on both axes at once, so it needs both formulas combined — One formula — the ordinary distance formula, with $(0,0)$ substituted — is all that is needed.
Between $P(6, 8)$ and $Q(9, 0)$, which point is farther from the origin?
$Q$ is farther, since it has the larger single coordinate
they are equally far from the origin
neither — the origin distance formula only applies to points on an axis
$P$ is farther
Check your answer
$Q$ is farther, since it has the larger single coordinate — A single large coordinate does not decide this — both coordinates combine, giving $10$ for $P$ against $9$ for $Q$.
they are equally far from the origin — The distances come out to $10$ and $9$ once computed — not equal.
neither — the origin distance formula only applies to points on an axis — The origin-distance formula works for any point in the plane — it is not restricted to points on an axis.
✓ $P$ is farther — (D) $O P = \sqrt{6^2 + 8^2} = 10$ and $O Q = 9$, so $P$ sits farther out.
Once we know every vertex of a triangle or a quadrilateral, the distance formula can name the figure. Nothing needs to be drawn, and nothing measured by eye.
Compute the length of every side. For a quadrilateral, compute both diagonals too. Equal side lengths point to an isosceles or an equilateral triangle, or to a rhombus. The pattern in the numbers alone reads off the shape.
A right angle needs one further check. Equal sides alone are never enough. Three side lengths confirm a right angle only through the converse of the Pythagoras theorem. The square of the longest side must equal the sum of the squares of the other two.
A square needs both conditions at once: all four sides equal, and both diagonals equal too. A rhombus can carry equal sides with unequal diagonals. Only a square carries both.
Two more names come off the same numbers, one condition at a time. Take the vertices in the order the question gives them, going round the figure, or the word OPPOSITE means nothing. If both pairs of opposite sides are equal, the figure is at least a PARALLELOGRAM. If the two diagonals are equal as well, it is a RECTANGLE. If all four sides are equal on top of that, it is a SQUARE.
A parallelogram whose diagonals differ has no right angle in it, and a parallelogram whose four sides are not all equal is not a rhombus.
Your turn: a quadrilateral has all sides $5$, one diagonal $8$ and the other $6$. Is it a square? (Answer: no, it is a rhombus, since a square needs both diagonals equal.)
Both triangles carry the same one-tick mark on their equal sides, but only one of the two has a right angle.
The rectangle’s two short sides carry one tick, its two long sides carry two ticks, and its diagonals carry three ticks.
Check yourself
Given only the three side lengths of a triangle, how can you confirm it has a right angle without measuring any angle?
check the converse Pythagoras theorem on the three sides
check whether any two of the three sides are equal in length
check whether the longest side is exactly twice the shortest
draw the triangle to scale and measure the angle with a protractor
Check your answer
✓ check the converse Pythagoras theorem on the three sides — (A) The converse Pythagoras check — shorter sides’ squares summing to the longest side’s square — confirms a right angle.
check whether any two of the three sides are equal in length — Equal sides signal an isosceles triangle — a right angle needs the converse Pythagoras check instead.
check whether the longest side is exactly twice the shortest — There is no rule that the longest side must double the shortest — the real test is the converse Pythagoras relation.
draw the triangle to scale and measure the angle with a protractor — A protractor is not needed — the converse Pythagoras relation confirms the right angle from the side lengths alone.
$A(0, 0)$, $B(4, 0)$, $C(0, 3)$. What type of triangle is $A B C$?
right-angled and scalene
isosceles, since two of its sides look similar in length
equilateral, since all three sides come from the same right triangle
impossible to classify without also knowing the angles at $B$ and $C$
Check your answer
✓ right-angled and scalene — (A) $A B = 4$, $A C = 3$, $B C = 5$; since $3^2 + 4^2 = 5^2$, the triangle is right-angled, with all sides unequal.
isosceles, since two of its sides look similar in length — $4$ and $3$ are close but not equal — isosceles needs two sides to match exactly, not merely look similar.
equilateral, since all three sides come from the same right triangle — $4$, $3$ and $5$ are three different lengths — an equilateral triangle needs all three sides equal.
impossible to classify without also knowing the angles at $B$ and $C$ — The right angle is already confirmed from the three side lengths — $3^2 + 4^2 = 5^2$ needs no further angle.
$A(0, 0)$, $B(4, 0)$, $C(4, 3)$, $D(0, 3)$. What shape do these four points form?
a rectangle
a rhombus, since opposite sides come out equal
a square, since both diagonals come out equal
not a special quadrilateral, since the four points look irregular
Check your answer
✓ a rectangle — (A) Opposite sides match ($4$ and $3$) and both diagonals are equal ($5$), but adjacent sides differ, giving a rectangle.
a rhombus, since opposite sides come out equal — A rhombus needs all four sides equal — here only opposite sides match, giving a rectangle instead.
a square, since both diagonals come out equal — Equal diagonals alone do not make a square — the sides here are $4$ and $3$, not all equal.
not a special quadrilateral, since the four points look irregular — The computed sides and diagonals meet the rectangle conditions exactly — this is not an irregular shape.
Why must you check the diagonals as well as the sides to tell a square apart from a rhombus?
because a rhombus never has any diagonals at all
because the sides of a square are always longer than the sides of a rhombus
equal diagonals confirm the right angles a square needs
because checking the diagonals is only needed for triangles, not quadrilaterals
Check your answer
because a rhombus never has any diagonals at all — A rhombus has two diagonals like any quadrilateral — the diagonal LENGTHS are what distinguish it from a square.
because the sides of a square are always longer than the sides of a rhombus — Side length does not distinguish them — a rhombus and a square can share the exact same side length.
✓ equal diagonals confirm the right angles a square needs — (C) Equal sides give a rhombus; equal diagonals on top of that confirm the right angles of a square.
because checking the diagonals is only needed for triangles, not quadrilaterals — The diagonal check applies to quadrilaterals precisely — it is what separates a square from a rhombus.
A point equidistant from two given points $A$ and $B$ sits at the same distance from each. That is an equation waiting to be written down.
We set the two distance expressions equal, then square both sides. Squaring removes the square root from each side. We expand the brackets, and the squared terms cancel each other out. What is left is a linear equation in the point’s own coordinates.
Carry the expansion out in full before you collect anything. Stop halfway, and a square root stays tangled in the working, with nothing gained.
Put numbers on it. Which points $(x, y)$ are the same distance from $A(1, 2)$ as from $B(5, 2)$? Set the two distances equal and square both sides at once.
The $x^2$ terms cancel, and so does the whole $(y - 2)^2$, which is why no $y$ survives. What is left is $8 x = 24$, so $x = 3$: every point on the one vertical line halfway between $A$ and $B$.
Your turn: point $P(x, y)$ is equidistant from $A(0, 0)$ and $B(4, 0)$. Write down the equation you would square, before solving anything. (Answer: $\sqrt{x^2 + y^2} = \sqrt{(x - 4)^2 + y^2}$, and squaring it leaves $x = 2$.)
Point P on the y-axis is joined to points A and B by two segments, both tick-marked to show they share one length.
Check yourself
A point equidistant from $A$ and $B$ is one whose
distance from $A$ plus its distance from $B$ equals a fixed value
coordinates are exactly the average of $A$’s and $B$’s coordinates
lies exactly halfway between $A$ and $B$ along the segment $A B$
distance from $A$ equals its distance from $B$
Check your answer
distance from $A$ plus its distance from $B$ equals a fixed value — A fixed sum of distances describes an ellipse — “equidistant” means the two distances are equal, not that they add to a constant.
coordinates are exactly the average of $A$’s and $B$’s coordinates — The average of $A$ and $B$’s coordinates gives only the midpoint of $A B$ — one equidistant point among infinitely many others.
lies exactly halfway between $A$ and $B$ along the segment $A B$ — An equidistant point need not lie on segment $A B$ at all — it can sit anywhere on the perpendicular bisector.
✓ distance from $A$ equals its distance from $B$ — (D) “Equidistant” means the distance to $A$ and the distance to $B$ come out equal.
Why does the method square both distance expressions before equating them, rather than equating the square roots directly?
it removes the square roots, leaving a simple equation
because square roots cannot be set equal to each other
because squaring makes both distances positive, which they were not before
because the two distances must first be converted to whole numbers
Check your answer
✓ it removes the square roots, leaving a simple equation — (A) Squaring both sides removes the square roots, leaving a simple equation to solve.
because square roots cannot be set equal to each other — Two square roots CAN be set equal directly — squaring first is just the easier route to solve from.
because squaring makes both distances positive, which they were not before — A distance is already non-negative before squaring — squaring here is about removing the roots, not fixing a sign.
because the two distances must first be converted to whole numbers — No conversion to whole numbers is needed — the method works directly with whatever the coordinates give.
Find a point on the $y$-axis equidistant from $C(5, 2)$ and $D(-1, -4)$.
$(0, -1)$
$(0, 5)$
$(0, 1)$
the midpoint of $C D$, at $(2, -1)$
Check your answer
$(0, -1)$ — Solving the equation carefully gives $y = 1$, not $-1$ — check the sign when expanding $(-4 - y)^2$.
$(0, 5)$ — $(0, 5)$ just copies $C$’s $y$-coordinate — the actual equidistant point comes from solving the equation, giving $(0, 1)$.
✓ $(0, 1)$ — (C) Setting $25 + (2 - y)^2 = 1 + (-4 - y)^2$ and solving gives $y = 1$, so the point is $(0, 1)$.
the midpoint of $C D$, at $(2, -1)$ — The midpoint of $C D$ is $(2, -1)$, which is not even on the $y$-axis — this question needs a point that is.
Why does squaring turn the equal-distance equation into something easy to solve, rather than a messy one?
the squared ($x^2$, $y^2$) terms cancel on both sides
because the coordinates of $A$ and $B$ are always small numbers
because one of the two points is always the origin
because the square root is dropped before the equation is even written
Check your answer
✓ the squared ($x^2$, $y^2$) terms cancel on both sides — (A) The $x^2$ and $y^2$ terms cancel out on both sides, leaving a simple linear equation.
because the coordinates of $A$ and $B$ are always small numbers — The size of the coordinates is irrelevant — the simplification comes from the quadratic terms cancelling, not from small numbers.
because one of the two points is always the origin — Neither point needs to be the origin — the quadratic terms cancel for any two points once both sides are expanded.
because the square root is dropped before the equation is even written — The square root is dropped by squaring — it is that squaring, not an earlier drop, that causes the quadratic terms to cancel.
The point $P$ divides the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m : n$. Its coordinates come from one formula: $((m x_2 + n x_1)/(m + n), (m y_2 + n y_1)/(m + n))$.
Look closely at which letter multiplies which point’s coordinates. *$m$ pairs with $B$’s coordinates, and $n$ pairs with $A$’s coordinates. That is $m x_2$, never $m x_1$.* We will use this exact pairing in every problem that follows. The ratio $m : n$ names $A$ first and $B$ second, but the formula’s own pairing runs the other way round.
Try it yourself on a ratio of $2 : 3$ dividing $A$ and $B$. It multiplies $B$’s coordinates by $2$, and $A$’s coordinates by $3$.
Your turn: a ratio $3 : 5$ divides the segment joining $A$ and $B$. Which point’s coordinates get multiplied by $3$? (Answer: $B$’s, since $m$ always pairs with the second point.)
Two right-triangle constructions turn the ratio’s numbers into actual leg lengths, once for a 1 to 2 division and once for the midpoint.
Check yourself
$P$ divides the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m : n$. $P$’s coordinates are
$(m x_2 + n x_1, m y_2 + n y_1)$, without dividing by $m + n$
$((x_1 + x_2)/(m + n), (y_1 + y_2)/(m + n))$, an unweighted sum
$((m x_2 + n x_1)/(m + n), (m y_2 + n y_1)/(m + n))$
Check your answer
$(m x_2 + n x_1, m y_2 + n y_1)$, without dividing by $m + n$ — This leaves out dividing by $m + n$ — without it, the result is not even a point between $A$ and $B$.
$((x_1 + x_2)/(m + n), (y_1 + y_2)/(m + n))$, an unweighted sum — This adds the coordinates without weighting either point by $m$ or $n$ — the ratio never enters the calculation.
$((x_2 - x_1)/(m + n), (y_2 - y_1)/(m + n))$, borrowing a subtraction — Subtracting the coordinates is the distance formula’s move — the section formula needs a weighted SUM instead.
✓ $((m x_2 + n x_1)/(m + n), (m y_2 + n y_1)/(m + n))$ — (D) Each coordinate is a weighted sum of the two points’ coordinates, divided by $m + n$.
The section formula’s coordinates are a weighted blend of $A$’s and $B$’s coordinates. What idea does that blend capture?
$P$ sits partway from $A$ to $B$, at a fraction set by $m$ and $n$
the blend is only a shortcut for finding the midpoint, whatever $m$ and $n$ are
the blend gives the point exactly halfway between $A$, $B$ and the origin
the blend gives the average of the two points’ distances from the origin
Check your answer
✓ $P$ sits partway from $A$ to $B$, at a fraction set by $m$ and $n$ — (A) The weights $m$ and $n$ set how far along the segment from $A$ to $B$ the point $P$ sits.
the blend is only a shortcut for finding the midpoint, whatever $m$ and $n$ are — Only $m = n$ gives the midpoint — other ratios weight the blend toward one end or the other.
the blend gives the point exactly halfway between $A$, $B$ and the origin — The origin plays no part in this formula — it blends only $A$ and $B$, weighted by the ratio.
the blend gives the average of the two points’ distances from the origin — The blend combines the two points’ coordinates directly — it says nothing about distance from the origin.
In the formula, $x_2$ (point $B$’s coordinate) is multiplied by $m$. Which part of the ratio $m : n$ does that $m$ measure?
the share of the segment counted from $B$’s side, that is, $P B$
the share of the segment counted from $A$’s side, that is, $A P$
the total length of the segment $A B$
the distance of $P$ from the origin
Check your answer
the share of the segment counted from $B$’s side, that is, $P B$ — $m$ pairs with $A P$, not $P B$ — pairing it with $P B$ instead is the swap this chapter warns against.
✓ the share of the segment counted from $A$’s side, that is, $A P$ — (B) $m$ is the $A P$ share of the ratio $A P : P B$, and it is the weight attached to $B$’s coordinate.
the total length of the segment $A B$ — $m$ is only one part of the ratio $A P : P B$ — it is not the full length of $A B$.
the distance of $P$ from the origin — The section formula never measures distance from the origin — $m$ measures the $A P$ share of the ratio.
Find the point dividing the segment joining $A(-1, 3)$ and $B(9, 8)$ internally in the ratio $2 : 3$.
$(5, 6)$
$(4, 5.5)$
$(17, 22)$
$(3, 5)$
Check your answer
$(5, 6)$ — $(5, 6)$ is the point for the swapped ratio $3 : 2$ — using $m = 2$, $n = 3$ correctly gives $(3, 5)$.
$(4, 5.5)$ — $(4, 5.5)$ is the midpoint of $A B$ (ratio $1 : 1$) — the question asks for ratio $2 : 3$, giving $(3, 5)$.
$(17, 22)$ — Adding the ratio numbers into the coordinates directly skips the weighted-average formula, which gives $(3, 5)$.
A student’s notes show the section formula as $((x_1 + x_2)/(m + n), (y_1 + y_2)/(m + n))$. What is wrong with this?
nothing — this is a correct, simplified version of the formula
the numerators are missing their weights
the denominator should be $2(m + n)$, not $m + n$
the formula should use $x_1 - x_2$ and $y_1 - y_2$ instead of a sum
Check your answer
nothing — this is a correct, simplified version of the formula — This is not a valid simplification — without the weights $m$ and $n$, the ratio never enters the formula at all.
✓ the numerators are missing their weights — (B) The numerators need the ratio weights, $m x_2 + n x_1$, not the unweighted sum $x_1 + x_2$.
the denominator should be $2(m + n)$, not $m + n$ — The denominator $m + n$ is correct — the missing piece is the weighting $m$ and $n$ in the numerator.
the formula should use $x_1 - x_2$ and $y_1 - y_2$ instead of a sum — Subtraction is the distance formula’s move — the section formula needs a weighted SUM of the two coordinates.
The parallelogram’s diagonals cross at one shared point, which is the midpoint of each.
The midpoint of a segment is nothing new. It is the section formula’s own special case, at the ratio $1 : 1$.
We set $m = n$ in the section formula. Both denominators become $m + n$, a number that cancels against itself either way. The midpoint works out to $((x_1 + x_2)/2, (y_1 + y_2)/2)$. Average the $x$’s, then average the $y$’s.
Try it on $(2, 4)$ and $(8, 10)$. You should get the midpoint $(5, 7)$, with no ratio bookkeeping required at all.
Your turn: find the midpoint of $(4, 2)$ and $(10, 8)$. (Answer: $(7, 5)$, since $4$ and $10$ average to $7$, and $2$ and $8$ average to $5$.)
Check yourself
The midpoint of the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is
$(x_1 + x_2, y_1 + y_2)$
$((x_1 + x_2)/(m + n), (y_1 + y_2)/(m + n))$
$((x_1 + x_2)/2, (y_1 + y_2)/2)$
Check your answer
$(x_1 + x_2, y_1 + y_2)$ — This leaves out dividing by $2$ — without it, the point is not the midpoint, or even between $A$ and $B$.
$((x_1 + x_2)/(m + n), (y_1 + y_2)/(m + n))$ — For the midpoint, $m = n = 1$, so $m + n = 2$ — the denominator should already be simplified to $2$.
✓ $((x_1 + x_2)/2, (y_1 + y_2)/2)$ — (C) The midpoint averages the two points’ coordinates, each divided by $2$.
Why is the midpoint formula the special case of the section formula where $m = n$?
because a midpoint always lies at the exact centre of the coordinate plane
equal weights $m = n$ collapse the blend into a plain average
because $m = n$ forces both points to have the same coordinates
because the section formula only works at all when $m = n$
Check your answer
because a midpoint always lies at the exact centre of the coordinate plane — A midpoint is the centre of segment $A B$, not the centre of the whole coordinate plane — that depends only on $A$ and $B$.
✓ equal weights $m = n$ collapse the blend into a plain average — (B) Equal weights $m = n$ collapse the section formula’s weighted blend into a plain average.
because $m = n$ forces both points to have the same coordinates — Equal weights $m = n$ say nothing about $A$ and $B$’s own coordinates — they only make the blend even between the two.
because the section formula only works at all when $m = n$ — The section formula works for any ratio $m : n$ — $m = n$ is only the one special case that gives the midpoint.
In parallelogram $A B C D$, $A(1, 2)$, $B(4, 3)$, $C(6, 7)$. The diagonals bisect each other. Find $D$.
Given a point already known to lie on a segment, we can find the ratio it divides the segment in.
Write the unknown ratio as $k : 1$, so only one unknown remains. Substitute the point’s $x$-coordinate into the section formula’s $x$-half. Solve the resulting equation for $k$.
*Checking the answer against the $y$-coordinate too is not optional.* Both coordinates must agree on the same value of $k$. A mismatch means an arithmetic slip was made somewhere earlier: go back and find it.
Your turn: point $(2, 0)$ divides $(0, 0)$ and $(6, 0)$ in what ratio? (Answer: $1 : 2$, since $(2, 0)$ is a third of the way from $(0, 0)$ to $(6, 0)$.)
A SCHOLAR INDIA REMEMBERS
You solved for $k$ from the $x$-coordinate. Now the $y$-coordinate gives a different $k$. Check the arithmetic once. If it is clean, the point was never on the segment. The second reading is telling you something. Do not argue with it. Correct the working, or correct the claim that the point lies on $A B$.
Brahmagupta · the brahminy kite
Check yourself
Given a point known to lie on segment $A B$, the standard method for finding the ratio it divides $A B$ in is to
measure the segment with a ruler on a scale drawing
always assume the ratio is $1 : 1$, since that is the most common case
write the ratio as $k : 1$ and solve for $k$
swap $m$ and $n$ from a ratio already used elsewhere in the chapter
Check your answer
measure the segment with a ruler on a scale drawing — No ruler or drawing is needed — the ratio comes from solving an equation built from the section formula.
always assume the ratio is $1 : 1$, since that is the most common case — Assuming $1 : 1$ in advance defeats the purpose — the ratio is found by solving for $k$, not guessed.
✓ write the ratio as $k : 1$ and solve for $k$ — (C) Writing the ratio as $k : 1$ turns two unknowns into one, solvable from the point’s coordinate.
swap $m$ and $n$ from a ratio already used elsewhere in the chapter — Reusing a ratio from a different question does not work — this point’s own coordinates must be substituted and solved.
Why does writing the ratio as $k : 1$ make it easier to solve for?
it leaves only one unknown, $k$, to solve for
it guarantees the ratio will always come out to a whole number
it changes the section formula into the midpoint formula
it removes the need to know the point’s coordinates at all
Check your answer
✓ it leaves only one unknown, $k$, to solve for — (A) Writing $m : n$ as $k : 1$ leaves only one unknown, $k$, to solve for.
it guarantees the ratio will always come out to a whole number — $k$ can come out as a fraction just as easily as a whole number — $k : 1$ simplifies the algebra, not the answer’s form.
it changes the section formula into the midpoint formula — Writing the ratio as $k : 1$ still uses the general section formula — it does not force the midpoint case $m = n$.
it removes the need to know the point’s coordinates at all — The point’s coordinates are essential here — they are substituted into the formula to solve for $k$.
Why might you need the $y$-coordinate equation instead of the $x$-coordinate one to find $k$?
because the $y$-coordinate equation is always more accurate than the $x$-coordinate one
on a vertical segment, the $x$-equation gives no information
because CBSE convention requires solving with $y$ whenever both are available
because the $x$-coordinate equation only works for points in the first quadrant
Check your answer
because the $y$-coordinate equation is always more accurate than the $x$-coordinate one — Neither equation is more accurate — the $x$-equation simply carries no information on a vertical segment.
✓ on a vertical segment, the $x$-equation gives no information — (B) On a vertical segment, the $x$-coordinate equation reduces to $0 = 0$, so the $y$-equation is needed.
because CBSE convention requires solving with $y$ whenever both are available — There is no such convention — either equation works fine as long as it is not the degenerate $0 = 0$ case.
because the $x$-coordinate equation only works for points in the first quadrant — The $x$-equation works in any quadrant — the real exception is a vertical segment, where it carries no information.
Point $(1, 3)$ lies on the segment joining $A(-3, 1)$ and $B(9, 7)$. Find the ratio it divides the segment in.
$2 : 1$
$1 : 2$
$1 : 3$
$3 : 1$
Check your answer
$2 : 1$ — $2 : 1$ is the reversed ratio — solving carefully for $k$ gives $1 : 2$, not its swap.
✓ $1 : 2$ — (B) Solving $(9k - 3)/(k + 1) = 1$ gives $k = 0.5$, so the ratio is $1 : 2$ — the $y$-coordinate check confirms it.
$1 : 3$ — Solving $(9k - 3)/(k + 1) = 1$ carefully gives $k = 0.5$, so the ratio is $1 : 2$, not $1 : 3$.
$3 : 1$ — This both mis-solves for $k$ and reverses the ratio — the correct value is $k = 0.5$, giving $1 : 2$.
Put $n$ with $x_2$ and $m$ with $x_1$ in the section formula. Surely the labels do not matter, since both numbers get used either way.
They do matter. Here is why. Take $A(0, 0)$ and $B(10, 0)$, divided in the ratio $1 : 4$. Here $m = 1$ and $n = 4$.
Done correctly, the formula gives $x = (m x_2 + n x_1)/(m + n)$. Substitute the numbers: $x = (1 \cdot 10 + 4 \cdot 0)/5 = 2$. That is the point $(2, 0)$, close to $A$, exactly as the smaller ratio number should place it.
Now swap the labels. Put $n$ where $m$ belongs, and $m$ where $n$ belongs. The formula gives $x = (4 \cdot 10 + 1 \cdot 0)/5 = 8$ instead, so the point becomes $(8, 0)$.
It is the correct point for the ratio $4 : 1$, not $1 : 4$. The two towns’ roles have been reversed.
One safety check catches the slip every time. Does the point sit nearer the label with the smaller ratio number? $(2, 0)$ sits near $A$, matching the smaller ratio number, $1$. $(8, 0)$ does not, and that alone tells us a swap happened somewhere in the working.
Your turn: $A(0, 0)$ and $B(20, 0)$ are divided in the ratio $1 : 3$. Is the dividing point $(5, 0)$ or $(15, 0)$? (Answer: $(5, 0)$, since $x = (1 \cdot 20 + 3 \cdot 0)/4 = 5$, near $A$.)
The point at 2, 0 sits near one end, and the swapped point at 8, 0 sits near the other end.
First try
A is $(0, 0)$ and B is $(10, 0)$, ratio $1 : 4$. I put the $4$ with $x_2$ and the $1$ with $x_1$, and got $x = (4 \cdot 10 + 1 \cdot 0)/5 = 8$.
Second look
The ratio’s first number multiplies the second point, B, and the second number multiplies the first point, A. So $m = 1$ pairs with $x_2 = 10$ and $n = 4$ pairs with $x_1 = 0$, giving $x = (1 \cdot 10 + 4 \cdot 0)/5 = 2$.
The ratio’s first number always goes with the second point, B. Keep that order every time.
Find the point dividing the segment from A to B in the ratio 1 to 4.
Weaker. Put $n$ with $x_2$ and $m$ with $x_1$. Surely it does not matter which label goes where.
For $A(0, 0)$ and $B(10, 0)$ in ratio $1 : 4$, that gives $x = (4 \cdot 10 + 1 \cdot 0)/5 = 8$, so the point comes out $(8, 0)$.
But $(8, 0)$ sits nearer $B$, not $A$. A point dividing in $1 : 4$ should sit nearer $A$, the end with the smaller number. Something has gone wrong.
Stronger. Put $m$ with $x_2$ and $n$ with $x_1$, matching the ratio’s own order. For $A(0, 0)$ and $B(10, 0)$ in ratio $1 : 4$, $m = 1$ and $n = 4$, so $x = (1 \cdot 10 + 4 \cdot 0)/5 = 2$, giving the point $(2, 0)$.
$(2, 0)$ sits nearer $A$, matching the smaller ratio number, $1$. This is the correct point for $1 : 4$.
Check yourself
$A(0, 0)$, $B(10, 0)$, ratio $1 : 4$. A student puts $n$ with $x_2$ and $m$ with $x_1$, and gets the point $(8, 0)$. What went wrong?
nothing went wrong — $(8, 0)$ is correct for the ratio $1 : 4$
$m$ and $n$ were swapped
the ratio should have been added instead of used as separate weights
the $y$-coordinates were mixed up instead of the $x$-coordinates
Check your answer
nothing went wrong — $(8, 0)$ is correct for the ratio $1 : 4$ — $(8, 0)$ is the point for ratio $4 : 1$, not $1 : 4$ — swapping $m$ and $n$ changed which ratio was actually computed.
✓ $m$ and $n$ were swapped — (B) Swapping $m$ and $n$ gives the point for the reversed ratio $4 : 1$; the ratio $1 : 4$ point is $(2, 0)$.
the ratio should have been added instead of used as separate weights — Adding the ratio numbers is not how the formula works — the real error here is that $m$ and $n$ were swapped.
the $y$-coordinates were mixed up instead of the $x$-coordinates — Both points share $y = 0$, so a $y$-coordinate mix-up cannot explain $(8, 0)$ — the actual error is the $m$/$n$ swap in the $x$-coordinate.
$A(2, 2)$, $B(12, 7)$, ratio $2 : 3$. A student computes the dividing point and gets $(8, 5)$. What went wrong, if anything?
nothing went wrong — $(8, 5)$ is correct
the midpoint formula was used by mistake instead of the section formula
$m$ and $n$ were swapped
the coordinates of $A$ and $B$ were subtracted instead of combined with the ratio
Check your answer
nothing went wrong — $(8, 5)$ is correct — $(6, 4)$ is the correct point for ratio $2 : 3$ — $(8, 5)$ comes from swapping $m$ and $n$.
the midpoint formula was used by mistake instead of the section formula — The midpoint of $A B$ is $(7, 4.5)$ — that does not match $(8, 5)$ either, so this is not a midpoint mix-up.
✓ $m$ and $n$ were swapped — (C) Using $m = 2$ with $x_2$ correctly gives $(6, 4)$; swapping to $m = 3$ gives the wrong point $(8, 5)$.
the coordinates of $A$ and $B$ were subtracted instead of combined with the ratio — Subtracting the coordinates would give a very different result, not $(8, 5)$ — the actual error is the $m$/$n$ swap.
Why does it matter which point is labelled $A$ and which is labelled $B$ when applying a ratio $m : n$?
swapping the labels swaps which end the ratio counts from
it does not matter — the formula is symmetric in $m$ and $n$
it only matters when one of the points has negative coordinates
it only matters when finding a midpoint
Check your answer
✓ swapping the labels swaps which end the ratio counts from — (A) $m : n$ means $A P : P B$; swapping the labels reverses which end the ratio counts from.
it does not matter — the formula is symmetric in $m$ and $n$ — The formula is NOT symmetric in $m$ and $n$ — swapping them gives the mirror-image point, unless $m = n$.
it only matters when one of the points has negative coordinates — The labelling matters regardless of sign — swapping $A$ and $B$ changes the ratio direction for any coordinates.
it only matters when finding a midpoint — It is the OPPOSITE — the midpoint ($m = n$) is the one case where the labelling stops mattering.
A worked solution divides $A(3, -2)$ and $B(11, 10)$ in ratio $3 : 1$ by writing $x = (1 \cdot 11 + 3 \cdot 3)/4$, $y = (1 \cdot 10 + 3 \cdot (-2))/4$. What is the correct fix?
leave the working as it is — it already applies the formula correctly
swap which point is called $A$ and which is called $B$, and keep the same weights
swap the weights back: $m$ with $x_2$, $n$ with $x_1$
average all four given numbers together as a rough estimate
Check your answer
leave the working as it is — it already applies the formula correctly — The working has $m$ and $n$ swapped — correcting the weights gives $(9, 7)$, not the point this working produces.
swap which point is called $A$ and which is called $B$, and keep the same weights — Swapping the point labels alone does not fix this — the weights $m$ and $n$ themselves need to go with the right coordinate.
average all four given numbers together as a rough estimate — Averaging four unrelated numbers gives no meaningful point — the exact fix is weighting $x_2$ by $m$ and $x_1$ by $n$.
For which ratio does swapping $m$ and $n$ leave the resulting point completely unchanged?
$2 : 1$
any ratio at all — swapping $m$ and $n$ never changes the point
no ratio — swapping $m$ and $n$ always changes the point
$1 : 1$, the midpoint
Check your answer
$2 : 1$ — Swapping $2 : 1$ to $1 : 2$ moves the point — only an EQUAL ratio, $1 : 1$, is immune to the swap.
any ratio at all — swapping $m$ and $n$ never changes the point — Swapping changes the point for every ratio except $1 : 1$ — this is exactly the mix-up the chapter warns against.
no ratio — swapping $m$ and $n$ always changes the point — $1 : 1$ IS immune — with $m = n$, swapping the two equal weights changes nothing.
✓ $1 : 1$, the midpoint — (D) Only $m = n$ (the ratio $1 : 1$) makes the formula symmetric, so swapping changes nothing.
$\sqrt{(6 - 1)^2 + (14 - 2)^2}$ Substitute the two points, $(1, 2)$ and $(6, 14)$, straight into the distance formula.
$= \sqrt{5^2 + 12^2} = \sqrt{25 + 144}$ Subtract inside each bracket, then square each result.
$= \sqrt{169} = 13$ Add the two squares, then take the square root.
$13^2 = 169$ and $5^2 + 12^2 = 25 + 144 = 169$ Check it by squaring the answer back. It matches the sum of the two squared legs you started from, exactly.
the distance between two points
Label the two points Call them $(x_1, y_1)$ and $(x_2, y_2)$. For $(1, 2)$ and $(6, 14)$, $x_1 = 1$, $y_1 = 2$, $x_2 = 6$, $y_2 = 14$.
Subtract, then square Find $x_2 - x_1$ and $y_2 - y_1$, then square each result. Here that is $5^2$ and $12^2$.
Add, then root Add the two squares, then take the square root. $25 + 144 = 169$, and $\sqrt{169} = 13$.
Check the answer Square the answer back. $13^2 = 169$, which matches the sum of the two squared legs.
$\sqrt{14}$ — Adding the legs directly ($8 + 6 = 14$) skips the squaring step — $\sqrt{8^2 + 6^2} = 10$ is the actual distance.
$14$ — Summing the legs plainly ($8 + 6$) is not the distance formula — squaring, adding, then rooting gives $10$.
$8$ — Using only the larger leg ($8$) ignores the vertical leg entirely — both legs combine to give $10$.
A worked line reads: $\sqrt{(6 - 1) + (14 - 2)} = \sqrt{17}$, for the distance between $(1, 2)$ and $(6, 14)$. What is the correct fix?
take the square root of each difference separately, then add the two roots
drop the square root entirely and just add the two differences
square each difference before adding: $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$
multiply the two differences together instead of adding them
Check your answer
take the square root of each difference separately, then add the two roots — Square roots do not add like this — the fix is to square each difference first, giving $\sqrt{5^2 + 12^2} = 13$.
drop the square root entirely and just add the two differences — Dropping the square root does not fix anything — the missing step is squaring the differences before adding them.
✓ square each difference before adding: $\sqrt{5^2 + 12^2} = \sqrt{169} = 13$ — (C) Each difference must be squared before adding: $\sqrt{5^2 + 12^2} = 13$, not $\sqrt{17}$.
multiply the two differences together instead of adding them — The formula adds the squared legs, not multiplies them — squaring first gives $\sqrt{5^2 + 12^2} = 13$.
Which of these correctly computes $\sqrt{5^2 + 12^2}$?
$A(1, 1)$, $B(4, 5)$, $C(8, 2)$, $D(5, -2)$ List the four vertices before computing anything.
$A B = \sqrt{3^2 + 4^2} = 5$ Apply the distance formula to the first side.
$B C = 5$, $C D = 5$, $D A = 5$ The same formula applies to the remaining three sides. Every one comes out equal too.
$A C = \sqrt{7^2 + 1^2} = 5 \sqrt{2}$, $B D = \sqrt{1^2 + 7^2} = 5 \sqrt{2}$ Compute both diagonals the same way. They too come out equal.
all four sides equal at $5$, both diagonals equal at $5 \sqrt{2}$ Equal sides alone would only confirm a rhombus. Equal diagonals as well are what confirm a square, since a parallelogram with equal diagonals has a right angle at every corner.
$A B^2 + B C^2 = 25 + 25 = 50 = (5 \sqrt{2})^2 = A C^2$ *Check the right angle at $B$ independently, using the Pythagoras converse.* The two squared sides add up to the squared diagonal, confirming the corner really is $90^\circ$.
the four sides
the two diagonals
Four pegs and lengths of string mark the quadrilateral Ananya is checking: the four sides, and both diagonals across it.
confirming a quadrilateral is a square
List the vertices Write down all four points before computing anything. Here that is $A(1, 1)$, $B(4, 5)$, $C(8, 2)$, $D(5, -2)$.
Measure every side Apply the distance formula to all four sides in turn. Each one here comes out to $5$.
Measure both diagonals Apply the distance formula to $A C$ and $B D$ too. Each one here comes out to $5 \sqrt{2}$.
Compare the lengths Equal sides alone would only confirm a rhombus. Equal diagonals as well point to a square.
Check the right angle Use the Pythagoras converse on two sides and a diagonal at one corner. Here $5^2 + 5^2 = 50 = (5 \sqrt{2})^2$, so the corner really is $90^\circ$.
Check yourself
$A(0, 0)$, $B(4, 3)$, $C(8, 0)$, $D(4, -3)$. What shape do these four points form?
a square, since all four sides come out equal
a rectangle, since opposite sides are parallel
not a special quadrilateral, since the diagonals are different lengths
a rhombus
Check your answer
a square, since all four sides come out equal — Equal sides alone give a rhombus — the diagonals here, $8$ and $6$, are unequal, so this is not a square.
a rectangle, since opposite sides are parallel — A rectangle also needs equal diagonals — here they are $8$ and $6$, unequal, so this is a rhombus instead.
not a special quadrilateral, since the diagonals are different lengths — Unequal diagonals only rule out a square — the four equal sides still make this shape a rhombus.
✓ a rhombus — (D) All four sides equal $5$; the diagonals, $8$ and $6$, are unequal, so this is a rhombus, not a square.
The worked example confirmed $A B C D$ was a square only after checking the diagonals, not just the sides. Why was that extra check necessary?
because the sides could have been measured incorrectly the first time
because a square must have more than four equal sides
because the diagonals must always be checked before the sides, never after
to confirm the right angles a square needs, beyond a rhombus’s equal sides
Check your answer
because the sides could have been measured incorrectly the first time — The diagonal check is not about catching a measuring mistake — it is what logically separates a rhombus from a square.
because a square must have more than four equal sides — A square has exactly four equal sides — the diagonal check is about confirming right angles, not adding more sides.
because the diagonals must always be checked before the sides, never after — The order does not matter — what matters is that BOTH the sides and the diagonals are checked before concluding square.
✓ to confirm the right angles a square needs, beyond a rhombus’s equal sides — (D) Equal sides give a rhombus; equal diagonals add the right-angle confirmation needed for a square.
A solution says: “all four sides of $A B C D$ came out equal, so $A B C D$ must be a square.” What is the flaw, and the correct fix?
no fix is needed — equal sides are enough to conclude square
the fix is to recompute the sides using the section formula instead of the distance formula
the fix is to check that the four points are listed in alphabetical order
equal sides give only a rhombus; also check the diagonals
Check your answer
no fix is needed — equal sides are enough to conclude square — Equal sides alone prove only a rhombus — confirming square needs the diagonals checked too.
the fix is to recompute the sides using the section formula instead of the distance formula — Side lengths are what the distance formula measures — the section formula finds a dividing point, not a length.
the fix is to check that the four points are listed in alphabetical order — The order the points are labelled in has no effect on the shape — the missing check is the diagonals, not the labelling.
✓ equal sides give only a rhombus; also check the diagonals — (D) The fix is to additionally check the diagonals — equal sides alone only confirm a rhombus.
The quadrilateral shows four sides tick-marked with one equal length, and two diagonals tick-marked with a second, longer length.
let the point be $(0, y)$ Any point on the $y$-axis has $x$-coordinate $0$.
$9 + (2 - y)^2 = 1 + (6 - y)^2$ Equate the squares of the distances to $A(3, 2)$ and $B(-1, 6)$ directly. This way, the square roots never appear.
$9 + 4 - 4 y + y^2 = 1 + 36 - 12 y + y^2$ Expand both squared brackets fully.
$y = 3$ The $y^2$ terms cancel on both sides, leaving a linear equation to solve.
the point is $(0, 3)$, distance $\sqrt{10}$ from each *Substitute $y = 3$ back in. Check that both distances really do agree.*
a point on an axis equidistant from two points
Set up the unknown point A point on the $y$-axis is always $(0, y)$. Here we want it equidistant from $A(3, 2)$ and $B(-1, 6)$.
Equate the squares Set the two squared distances equal, so no square root appears. Here that is $9 + (2 - y)^2 = 1 + (6 - y)^2$.
Expand fully Expand both brackets before collecting anything. $9 + 4 - 4 y + y^2 = 1 + 36 - 12 y + y^2$.
Cancel and solve The $y^2$ terms cancel, leaving one linear equation. Solving it gives $y = 3$.
Check both distances Substitute $y = 3$ back into both distances. Both come out to $\sqrt{10}$, so the point $(0, 3)$ checks out.
Check yourself
Find a point on the $x$-axis equidistant from $E(1, 5)$ and $F(4, -2)$.
$(4, 0)$
$(1, 0)$
the midpoint of $E F$, at $(2.5, 1.5)$
$(-4, 0)$
Check your answer
✓ $(4, 0)$ — (A) Setting $(x - 1)^2 + 25 = (x - 4)^2 + 4$ and solving gives $x = 4$, so the point is $(4, 0)$.
$(1, 0)$ — $(1, 0)$ just copies $E$’s $x$-coordinate — solving the actual equation gives $(4, 0)$.
the midpoint of $E F$, at $(2.5, 1.5)$ — The midpoint of $E F$ is $(2.5, 1.5)$, which is not on the $x$-axis — the question needs a point that is.
$(-4, 0)$ — Solving carefully gives $x = 4$, not $-4$ — check the sign when expanding $(x - 4)^2$.
The worked example finds ONE point on an axis equidistant from $A$ and $B$. Is that the only point in the plane equidistant from $A$ and $B$?
yes — it is the only equidistant point that exists
no — the whole perpendicular bisector is equidistant
yes, because equidistant points can only occur on one of the two axes
no, but only two such points exist in total
Check your answer
yes — it is the only equidistant point that exists — An entire line — the perpendicular bisector of $A B$ — is equidistant from $A$ and $B$, not just this one point.
✓ no — the whole perpendicular bisector is equidistant — (B) Infinitely many points on the perpendicular bisector of $A B$ are equidistant; the worked point is the one on the axis.
yes, because equidistant points can only occur on one of the two axes — Equidistant points are not restricted to the axes — they fill the whole perpendicular bisector line.
no, but only two such points exist in total — The perpendicular bisector is an entire line, giving infinitely many equidistant points, not just two.
A solution finds a point on the $y$-axis equidistant from $A(3, 2)$ and $B(-1, 6)$ by writing $9 + (2 - y)^2 = 1 + (6 + y)^2$ (a sign slip on the last term). What is the correct fix?
leave the equation as it is — a sign difference does not affect the final answer
start over by using $B$’s coordinates in place of $A$’s throughout
fix the sign: use $(6 - y)^2$, then solve to $y = 3$
multiply both sides by $-1$ to cancel the sign difference
Check your answer
leave the equation as it is — a sign difference does not affect the final answer — A sign difference inside a squared term does change the expanded equation — fixing it is what gives the correct $y = 3$.
start over by using $B$’s coordinates in place of $A$’s throughout — Swapping the points does not fix the sign error — the fix is correcting $(6 + y)^2$ to $(6 - y)^2$.
✓ fix the sign: use $(6 - y)^2$, then solve to $y = 3$ — (C) Correcting the sign to $(6 - y)^2$ and solving gives $y = 3$, matching the worked example’s point $(0, 3)$.
multiply both sides by $-1$ to cancel the sign difference — Multiplying the whole equation by $-1$ does not fix an error inside one term — the fix is correcting that one term’s sign.
$m = 1$, $n = 2$ Read the ratio $1 : 2$ directly as $m$ and $n$, for $A(2, -2)$ and $B(8, 10)$.
$x = (1 \cdot 8 + 2 \cdot 2)/3 = 4$ Substitute into the section formula’s $x$-half. $m$ multiplies $B$’s $x$-coordinate, and $n$ multiplies $A$’s.
$y = (1 \cdot 10 + 2 \cdot (-2))/3 = 2$ The same substitution on the $y$-half.
the point is $(4, 2)$ Pair the two coordinates you just found.
$A P = \sqrt{20} = 2 \sqrt{5}$ and $P B = \sqrt{80} = 4 \sqrt{5}$ Check the ratio by measuring both pieces with the distance formula. They come out $2 \sqrt{5} : 4 \sqrt{5} = 1 : 2$, exactly as given.
dividing a segment in a given ratio
Read off m and n Match the ratio’s two numbers to $m$ and $n$ in order. Here the ratio $1 : 2$ gives $m = 1$, $n = 2$, for $A(2, -2)$ and $B(8, 10)$.
Substitute into the x-half $m$ multiplies $B$’s $x$-coordinate, $n$ multiplies $A$’s. Here $x = (1 \cdot 8 + 2 \cdot 2)/3 = 4$.
Substitute into the y-half Use the same pairing on the $y$-coordinates. Here $y = (1 \cdot 10 + 2 \cdot (-2))/3 = 2$.
Pair and check The point is $(4, 2)$. Check it by measuring both pieces with the distance formula, and confirm the ratio.
Check yourself
Find the point dividing the segment joining $A(-4, -1)$ and $B(6, 9)$ internally in the ratio $3 : 2$.
$(0, 3)$
$(1, 4)$
$(2, 5)$
$(-2, 3)$
Check your answer
$(0, 3)$ — The weighted sums must be divided by $m + n = 5$ — skipping that step gives $(0, 3)$ instead of $(2, 5)$.
$(1, 4)$ — $(1, 4)$ is the midpoint of $A B$ — the question asks for ratio $3 : 2$, which gives $(2, 5)$.
In the worked example, $A(2, -2)$, $B(8, 10)$, ratio $1 : 2$ gives the point $(4, 2)$, which sits closer to $A$ than to $B$. Why does that make sense?
$A P:P B = 1:2$ makes $A P$ the smaller share, nearer $A$
it does not make sense — the point should sit closer to $B$ instead
it is just a coincidence for this particular pair of points
it only makes sense because $A$ has negative coordinates
Check your answer
✓ $A P:P B = 1:2$ makes $A P$ the smaller share, nearer $A$ — (A) $A P : P B = 1 : 2$ makes $A P$ the smaller share, so the point sits closer to $A$.
it does not make sense — the point should sit closer to $B$ instead — A ratio $1 : 2$ places the point nearer $A$, the smaller-share end — not nearer $B$.
it is just a coincidence for this particular pair of points — This is not a coincidence — any ratio $1 : 2$ places the dividing point nearer the smaller-share end, $A$.
it only makes sense because $A$ has negative coordinates — The sign of $A$’s coordinates is irrelevant here — closeness to $A$ comes from the ratio $1 : 2$ itself.
A solution dividing $A(2, -2)$, $B(8, 10)$ in ratio $1 : 2$ writes $x = (2 \cdot 8 + 1 \cdot 2)/3$ (the weights swapped). What is the correct fix?
leave the working as it is — both weightings give the same point anyway
swap the weights back: $m$ with $x_2$, $n$ with $x_1$
swap $A$ and $B$’s coordinates instead of the weights
divide by $m - n$ instead of $m + n$
Check your answer
leave the working as it is — both weightings give the same point anyway — The two weightings give different points here, since $1 n e 2$ — only the correctly weighted version gives the worked example’s answer.
✓ swap the weights back: $m$ with $x_2$, $n$ with $x_1$ — (B) Correct weighting gives $x = (1 \cdot 8 + 2 \cdot 2)/3 = 4$, matching the worked example’s point.
swap $A$ and $B$’s coordinates instead of the weights — Swapping the coordinates instead of the weights does not fix this — the weights $m$ and $n$ need to attach to the right point.
divide by $m - n$ instead of $m + n$ — The denominator is always $m + n$ — changing it to $m - n$ does not correct the swapped weights.
Here the point is the given and the ratio is the unknown, the reverse of the example before it.
Worked example
Finding an unknown ratio
write the ratio as $k : 1$ One unknown ratio needs only one unknown, for $A(-1, 7)$, $B(5, -5)$ and the known point $(1, 3)$.
$1 = (5 k - 1)/(k + 1)$ Substitute the known point’s $x$-coordinate into the section formula’s $x$-half.
$5 k - 1 = k + 1$, so $k = 1/2$ Clear the denominator, then collect terms and solve.
the ratio is $1 : 2$ $k = 1/2$ written as a ratio to $1$.
checked against the $y$-coordinate: $y = (1/2 \cdot (-5) + 1 \cdot 7)/(1/2 + 1) = 3$ *The same $k$ must satisfy both coordinates.* This confirms no arithmetic slip was made.
finding the ratio a point divides a segment in
Write the ratio Write the unknown ratio as $k : 1$. One unknown needs only one unknown, for $A(-1, 7)$, $B(5, -5)$ and the known point $(1, 3)$.
Substitute the x-coordinate Put the known point’s $x$-coordinate into the section formula’s $x$-half. Here $1 = (5 k - 1)/(k + 1)$.
Solve for k Clear the denominator, then collect terms. Here $5 k - 1 = k + 1$, so $k = 1/2$.
State the ratio Turn $k$ back into a ratio to $1$. Here $k = 1/2$ gives the ratio $1 : 2$.
Check with the other coordinate The same $k$ must satisfy the $y$-coordinate too. Here it does, confirming no arithmetic slip was made.
Check yourself
Point $(2, -3)$ lies on the segment joining $A(-4, 3)$ and $B(5, -6)$. Find the ratio it divides the segment in.
$1 : 2$
$2 : 3$
$2 : 1$
$3 : 2$
Check your answer
$1 : 2$ — $1 : 2$ is the reversed ratio — solving carefully for $k$ gives $2 : 1$, not its swap.
$2 : 3$ — Solving $(5k - 4)/(k + 1) = 2$ carefully gives $k = 2$, so the ratio is $2 : 1$, not $2 : 3$.
✓ $2 : 1$ — (C) Solving $(5k - 4)/(k + 1) = 2$ gives $k = 2$, confirmed by the $y$-coordinate, so the ratio is $2 : 1$.
$3 : 2$ — This both mis-solves for $k$ and reverses the ratio — the correct value is $k = 2$, giving $2 : 1$.
The worked example checks the ratio using both the $x$-coordinate equation AND the $y$-coordinate equation. Why check both, instead of stopping after the first?
because the $x$-coordinate equation is never reliable on its own
because CBSE marking always requires two separate equations to be shown
it confirms the point truly lies on the segment, not by coincidence
because the $y$-coordinate always gives a different value of $k$ than the $x$-coordinate does
Check your answer
because the $x$-coordinate equation is never reliable on its own — The $x$-coordinate equation is reliable on its own — checking $y$ too is confirmation, not a fix for something unreliable.
because CBSE marking always requires two separate equations to be shown — The reason is mathematical, not procedural — the second check confirms the point actually lies on the segment.
✓ it confirms the point truly lies on the segment, not by coincidence — (C) The second coordinate check confirms the point truly lies on the segment, ruling out a coincidence.
because the $y$-coordinate always gives a different value of $k$ than the $x$-coordinate does — For a point that truly lies on the segment, both coordinates give the SAME $k$ — that agreement is the point of checking both.
A solution finds the ratio for point $(1, 3)$ on segment $A(-3, 1)$, $B(9, 7)$ by writing $(9k - 3)/(k + 1) = 1$ and solving to $k = 2$. What is the correct fix?
keep $k = 2$ — it is already correct
re-solve: $9k - 3 = k + 1$ gives $k = 0.5$, ratio $1 : 2$
swap $A$ and $B$ in the formula and keep $k = 2$
use the $y$-coordinate equation instead, and discard the $x$-coordinate one
Check your answer
keep $k = 2$ — it is already correct — Substituting $k = 2$ back into $(9k - 3)/(k + 1) = 1$ does not check out — solving carefully gives $k = 0.5$ instead.
✓ re-solve: $9k - 3 = k + 1$ gives $k = 0.5$, ratio $1 : 2$ — (B) Correctly solving $9k - 3 = k + 1$ gives $8k = 4$, so $k = 0.5$, matching ratio $1 : 2$.
swap $A$ and $B$ in the formula and keep $k = 2$ — Swapping $A$ and $B$ does not fix an algebra error — the fix is re-solving the equation correctly for $k$.
use the $y$-coordinate equation instead, and discard the $x$-coordinate one — The $x$-coordinate equation itself is fine — the error is in the algebra used to solve it, which gives $k = 0.5$.
DISTANCE: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For $(1, 2)$ and $(6, 14)$, $d = 13$.
FROM ORIGIN: $d = \sqrt{x^2 + y^2}$. For $(6, 8)$, $d = 10$.
SECTION: divide $(2, -2)$ and $(8, 10)$ in ratio $1 : 2$, and we land at $(4, 2)$.
MIDPOINT: the ratio $1 : 1$ case. Average $(2, 4)$ and $(8, 10)$, and we get $(5, 7)$.
LABEL ORDER: $m$ pairs with the second point, and $n$ with the first. Swap them, and $(2, 0)$ becomes $(8, 0)$: nearer $B$, when the smaller ratio number should keep it nearer $A$.
Look back over these five lines. Only two formulas are doing all the work.
The section formula works the other way round. Given a ratio, it finds the point. Given the point already found, the same formula finds the ratio back. The midpoint is its simplest case, the ratio $1 : 1$.
Your turn: find the midpoint of $(0, 2)$ and $(8, 4)$, then its distance from the origin. (Answer: midpoint $(4, 3)$, distance $5$, since $\sqrt{4^2 + 3^2} = \sqrt{25} = 5$.)
And Kabir’s question from the opening page. The two discs sit at $(3, 2)$ and $(7, 6)$, so one is $4$ tiles across and $4$ tiles up from the other, and the distance between them is $\sqrt{4^2 + 4^2} = \sqrt{32} = 4 \sqrt{2}$, about $5.7$ tiles. Four numbers in, one length out.
Check yourself
Two mobile towers stand at known coordinates, and an engineer wants to know how far apart they are. Which formula applies?
the section formula
the midpoint formula
the distance formula
no coordinate formula — this needs calculus instead
Check your answer
the section formula — The section formula locates a point from a ratio — a plain distance between two points needs the distance formula.
the midpoint formula — The midpoint formula gives a halfway point, not a length — the distance formula gives the actual gap between the towers.
✓ the distance formula — (C) A single length between two known points is exactly what the distance formula computes.
no coordinate formula — this needs calculus instead — No calculus is needed here — the distance formula, using only the two towers’ coordinates, answers this directly.
A road planner wants a rest stop $2/5$ of the way from town $A$ to town $B$ along a straight road. Which formula locates it?
the distance formula
the midpoint formula
the section formula, using the ratio $2 : 3$
the average of the two towns’ distances from the origin
Check your answer
the distance formula — The distance formula gives a length, not a new point — locating the rest stop needs the section formula.
the midpoint formula — The midpoint formula only finds the exact halfway point — a $2/5$ split needs the general section formula instead.
✓ the section formula, using the ratio $2 : 3$ — (C) $2/5$ of the way from $A$ to $B$ is the ratio $2 : 3$, which the section formula locates directly.
the average of the two towns’ distances from the origin — Distance from the origin has nothing to do with locating a point between the two towns — the section formula does.
Every problem in this chapter reduces to one of two formulas. What are they, and how does the midpoint fit in?
the distance formula and the midpoint formula, with the section formula as a special case of the midpoint
the section formula and the ratio-finding method, with distance as a special case of ratio-finding
just one formula, the section formula, since distance can always be recovered from it
distance (a length) and section (a point from a ratio); the midpoint is section’s $1:1$ case
Check your answer
the distance formula and the midpoint formula, with the section formula as a special case of the midpoint — This reverses the relationship — the midpoint is the SPECIAL CASE of the general section formula, not the reverse.
the section formula and the ratio-finding method, with distance as a special case of ratio-finding — Distance is its own separate formula — it is not a special case of ratio-finding, which instead runs the section formula in reverse.
just one formula, the section formula, since distance can always be recovered from it — The section formula locates a point — it does not by itself produce a length, which is what the separate distance formula is for.
✓ distance (a length) and section (a point from a ratio); the midpoint is section’s $1:1$ case — (D) The distance formula gives a length; the section formula gives a point from a ratio, with the midpoint as its $1 : 1$ case.
Check yourself: the whole chapter
According to this chapter’s opening claim, the distance between the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by which expression?
$(x_2 - x_1) + (y_2 - y_1)$
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2}$
$(x_2 - x_1)^2 + (y_2 - y_1)^2$
Check your answer
$(x_2 - x_1) + (y_2 - y_1)$ — A plain sum of differences is not a distance formula at all — the formula needs each difference SQUARED, then added, then a square root taken.
✓ $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ — (B) The chapter’s frame states the distance as $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ — subtract, square, add, then take the square root.
$\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2}$ — The formula subtracts the two x-coordinates and the two y-coordinates first, $x_2 - x_1$ and $y_2 - y_1$ — adding them instead gives a completely different, incorrect expression.
$(x_2 - x_1)^2 + (y_2 - y_1)^2$ — This is the correct sum of squares under the root, but the root itself is missing — the distance is the SQUARE ROOT of this sum, not the sum alone.
What is the distance between the points $(3, 2)$ and $(7, 5)$?
7
$\sqrt{74}$
5
$\sqrt{7}$
Check your answer
7 — 4 + 3 = 7 skips the formula entirely — the differences must be squared before they are added, and a square root taken at the end.
$\sqrt{74}$ — $\sqrt{7^2 + 5^2}$ finds the distance of $(7, 5)$ from the ORIGIN, not the distance between $(3, 2)$ and $(7, 5)$ — the two points’ coordinates must be subtracted first.
$\sqrt{7}$ — $\sqrt{4 + 3}$ adds the differences before squaring them — each difference must be squared FIRST, $4^2 = 16$ and $3^2 = 9$, then added, giving $\sqrt{25} = 5$.
Point $A(1, 1)$ and point $B(7, 13)$ are joined by a segment. Find the point that divides it, internally, in the ratio $1 : 2$.
$(5, 9)$
$(4, 7)$
$(4.5, 7.5)$
$(3, 5)$
Check your answer
$(5, 9)$ — Matching m = 1 to B’s coordinates and n = 2 to A’s gives (3, 5) — swapping the two finds the point for the reversed ratio, 2 : 1, instead.
$(4, 7)$ — (4, 7) is the plain midpoint of A and B, the ratio 1 : 1 case — the question asks for ratio 1 : 2, which is not the midpoint.
$(4.5, 7.5)$ — The denominator is m + n = 1 + 2 = 3, not a fixed 2 — using 2 regardless of the ratio gives the wrong point.
$(3, 4)$ — The midpoint averages the two points, which means ADDING their coordinates, not subtracting — (10 - 4)/2 = 3 is not part of the midpoint formula.
$(7, -4)$ — Both coordinates need dividing by 2, not just the x-coordinate — the y-sum -4 still needs halving to -2.
Point $(4, 4)$ lies on the segment joining $A(1, 2)$ and $B(10, 8)$. In what ratio, measured from $A$ to $B$, does it divide the segment?
$2 : 1$
$1 : 1$
$1 : 2$
$3 : 1$
Check your answer
$2 : 1$ — Writing the ratio as k : 1 and solving from A’s coordinate gives k = 1/2, a ratio of 1 : 2 from A to B — reporting 2 : 1 describes the reversed direction, from B to A.
$1 : 1$ — Not every point between A and B is the midpoint — solving the section formula for this exact point gives k = 1/2, a ratio of 1 : 2, not the midpoint’s 1 : 1.
✓ $1 : 2$ — (C) Writing the ratio as $k : 1$, $(k \cdot 10 + 1)/(k + 1) = 4$ gives $k = 1/2$, checked against the y-coordinate too — the ratio is $1 : 2$.
$3 : 1$ — Substituting (4, 4) into the section formula and solving for k gives k = 1/2, a ratio of 1 : 2 — 3 : 1 does not satisfy either coordinate’s equation.
You will use coordinates whenever a length or a matching point comes from two locations on a grid. Here are seven of those places.
Marking two points of a rangoli design. A rangoli design has two marked points, $A(3, 4)$ and $B(9, 12)$, on graph paper. The maths: the distance formula turns two points into one length. It uses the horizontal and vertical gap between them. The horizontal gap is $9 - 3 = 6$. The vertical gap is $12 - 4 = 8$, so the distance is $\sqrt{6^2 + 8^2} = 10$.
Two points on a grid are the ends of a right triangle whose legs are the horizontal and vertical gaps.
Point P divides the 15 m segment from A to B in the ratio one to two, landing 5 m from A.
Checking a patchwork block is square. A quilter marks four corners of a patchwork block on grid paper: $A(0, 0)$, $B(5, 12)$, $C(17, 7)$, $D(12, -5)$. The maths: compute every side and both diagonals with the distance formula. Equal sides and equal diagonals confirm a square. Every side comes out to $13$. Both diagonals come out to $13 \sqrt{2}$, so all four corners form a square.
Reading a savings tracker’s halfway point. A savings tracker plots two check-ins as points. $P(4, 800)$ and $Q(10, 2000)$ show week against balance in rupees. The maths: the midpoint formula averages the two coordinates, to estimate the balance halfway between the two check-ins. The midpoint is $((4 + 10)/2, (800 + 2000)/2) = (7, 1400)$, around week $7$ at about ₹$1400$.
Placing a practice cone at a fixed ratio. A coach marks two cones, $A(0, 0)$ and $B(9, 12)$, $15$ m apart on the field. The maths: the section formula finds a point that divides a segment in a given ratio, without measuring anything by hand. Dividing $A B$ in ratio $1 : 2$ from $A$ gives $((1 \cdot 9 + 2 \cdot 0)/3, (1 \cdot 12 + 2 \cdot 0)/3) = (3, 4)$, a new cone $5$ m from $A$.
Finding a bus stop’s distance from the centre. A town map puts the town centre at the origin, $O(0, 0)$, one grid unit to $100$ m. A new bus stop sits at $S(8, 15)$. The maths: the distance from the origin is the same distance formula, with one point fixed at the origin. The distance is $\sqrt{8^2 + 15^2} = 17$ units, which is $1.7$ km.
Placing a cutting table equally far from two doors. A tailor’s shop has two doors, $A(5, 4)$ and $B(-3, 8)$. The back wall is the line $x = 0$. A cutting table must sit on it, equally far from both doors. The maths: a point equidistant from two points makes their squared distances equal. Solving for the unknown coordinate finds the point. Squaring both distances and solving gives $y = 4$, so the table sits at $(0, 4)$, distance $5$ from each door.
Placing a water station between two stalls. An event planner’s spreadsheet marks two food stalls, $A(2, 3)$ and $B(17, 18)$. A water station must divide $A B$ in the ratio $2 : 3$. The maths: the spreadsheet must pair each ratio number with the correct point. Swapping the two numbers sends the station to the wrong spot. Correctly, the station lands at $(8, 9)$. Swapped, it lands at $(11, 12)$, closer to $B$ instead of $A$.
Your turn. A garden path has two lamp posts, $A(2, 3)$ and $B(10, 9)$. Find the distance between them, and the point exactly halfway. Answer: The distance is $\sqrt{8^2 + 6^2} = 10$ m. The halfway point, the midpoint, is $((2 + 10)/2, (3 + 9)/2) = (6, 6)$.
practice Find the distance between $(3, 1)$ and $(7, 4)$. (Worked in full below — read it, then do the next two the same way.)
practice Find the distance of the point $(-9, 40)$ from the origin. (Start the same way — the origin is the point $(0, 0)$, so put that in as the second point.)
practice Show that $A(1, 1)$, $B(5, 1)$ and $C(3, 5)$ are the vertices of an isosceles triangle. (Find all three side lengths with the same formula, then compare them.)
practice Find the distance between $(2, 3)$ and $(4, 1)$.
practice Find the distance of the point $(-5, 12)$ from the origin.
practice Two towns sit at $(0, 0)$ and $(24, 7)$ on a district grid, one grid unit to a kilometre. Find the straight-line distance between them.
practice Show that $(3, 0)$, $(6, 4)$ and $(-1, 3)$ are the vertices of an isosceles right triangle.
practice Show that $(1, 2)$, $(4, 2)$, $(4, 6)$ and $(1, 6)$ are the vertices of a rectangle.
practice Find a point on the $x$-axis equidistant from $(7, 6)$ and $(-3, 4)$.
$A B = 4$, $B C = \sqrt{20}$ and $C A = \sqrt{20}$. Two sides are equal, so the triangle is isosceles.
$\sqrt{2^2 + 2^2} = 2 \sqrt{2}$.
$\sqrt{(-5)^2 + 12^2} = 13$.
$\sqrt{24^2 + 7^2} = 25$ km.
The three sides are $5$, $5$ and $5 \sqrt{2}$; $5^2 + 5^2 = (5 \sqrt{2})^2$, confirming a right angle where the two equal sides meet.
Opposite sides come out $3$ and $4$, and both diagonals come out $5$, confirming a rectangle.
$(3, 0)$.
Exercise 7.1 — further practice
practice Find the distance between $(-5, 7)$ and $(-1, 4)$.
practice Which of these is the distance of the point $(8, 15)$ from the origin?
$17$
$23$
$\sqrt{161}$
$15$
practice Two water tanks on a farm map sit at $(0, 0)$ and $(9, 12)$, one grid unit to a metre. Find the distance between them.
practice Show that $A(0, 0)$, $B(4, 0)$ and $C(2, 5)$ are the vertices of an isosceles triangle.
practice Show that $A(4, -2)$, $B(7, 1)$ and $C(13, 7)$ are collinear.
practice Show that $A(0, 0)$, $B(3, 4)$, $C(0, 8)$ and $D(-3, 4)$, taken in order, are the vertices of a rhombus but not a square.
practice Show that $A(0, 0)$, $B(4, 0)$, $C(6, 3)$ and $D(2, 3)$, taken in order, are the vertices of a parallelogram that is not a rhombus.
practice Find a point on the $y$-axis equidistant from $A(3, -2)$ and $B(-5, 6)$.
practice Find the values of $y$ for which the distance between $(1, 2)$ and $(4, y)$ is $5$.
practice Find $k$ such that the point $(k, 3)$ is equidistant from $A(5, -2)$ and $B(1, 4)$.
Answers
$5$
A — $17$.
$15$ m.
$A C = B C = \sqrt{29}$ and $A B = 4$; two sides are equal, so the triangle is isosceles.
$A B = 3 \sqrt{2}$, $B C = 6 \sqrt{2}$ and $A C = 9 \sqrt{2}$; since $A B + B C = A C$, the three points are collinear.
Every side works out to $5$, but the diagonals $A C = 8$ and $B D = 6$ are not equal; equal sides with unequal diagonals give a rhombus, not a square.
$A B = C D = 4$ and $B C = D A = \sqrt{13}$; opposite sides are equal, giving a parallelogram, but all four sides are not equal, so it is not a rhombus.
practice Find the point that divides the segment joining $(1, 2)$ and $(7, 5)$ internally in the ratio $2 : 1$. (Worked in full below — read it, then do the next two the same way.)
practice Find the midpoint of the segment joining $(-3, 8)$ and $(5, -2)$. (The midpoint is the section formula with the two parts equal, so the halves cancel to a simple average.)
practice In what ratio does the point $(3, 0)$ divide the segment joining $(1, -2)$ and $(4, 1)$? (Start the same way, but here the point is known and the ratio is not — write the unknown ratio as $k : 1$, the way the chapter does, so only one unknown is left.)
practice Find the coordinates of the point dividing the segment joining $(4, -1)$ and $(-2, -3)$ internally in the ratio $1 : 2$.
practice Find the midpoint of the segment joining $(6, -5)$ and $(-2, 11)$.
practice Three vertices of a parallelogram $P Q R S$ are $P(1, 2)$, $Q(4, 3)$ and $R(6, 7)$. Find the fourth vertex $S$, using the fact that a parallelogram’s diagonals share their midpoint.
practice The point $(0, y)$ divides the segment joining $(-4, 6)$ and $(6, -4)$ internally. Find $y$ and the ratio of division.
Answers
$P = (5, 4)$.
$(1, 3)$.
$2 : 1$. Write the unknown ratio as $k : 1$, so only one unknown is left. The $x$ coordinate gives $3 = (4 k + 1) / (k + 1)$, so $3 k + 3 = 4 k + 1$ and $k = 2$, which is the ratio $2 : 1$. Check it against the $y$ coordinate: $(2 \cdot 1 + 1 \cdot (-2)) / (2 + 1) = 0$, which agrees.
$(2, -5/3)$.
$(2, 3)$.
$S(3, 6)$, from equating the midpoint of $P R$ with the midpoint of $Q S$.
Ratio $2 : 3$, $y = 2$.
Exercise 7.2 — further practice
practice Find the point dividing the segment joining $A(-4, 3)$ and $B(6, -2)$ internally in the ratio $3 : 2$.
practice Find the midpoint of the segment joining $(-3, 5)$ and $(9, -1)$.
practice Which of these is the point dividing the segment joining $(0, 0)$ and $(10, 15)$ internally in the ratio $2 : 3$?
$(4, 6)$
$(6, 9)$
$(2, 3)$
$(8, 12)$
practice Which of these is the midpoint of the segment joining $(-6, 3)$ and $(2, -9)$?
$(-2, -3)$
$(-4, 6)$
$(4, -3)$
$(-2, 3)$
practice Find the coordinates of the points of trisection of the segment joining $A(2, -4)$ and $B(-4, 8)$.
practice Three vertices of a parallelogram $P Q R S$, taken in order, are $P(1, -2)$, $Q(4, 0)$ and $R(7, 6)$. Find the fourth vertex $S$, using the fact that a parallelogram’s diagonals share their midpoint.
practice In what ratio does the $x$-axis divide the segment joining $A(1, 4)$ and $B(6, -6)$? Also find the point of division.
practice Find the ratio in which the point $(0, -1)$ divides the segment joining $A(3, 5)$ and $B(-2, -5)$.
practice $P$ lies on the segment joining $A(1, -2)$ and $B(9, 6)$ such that $A P = 3/8 \cdot A B$. Find the coordinates of $P$.
practice The point $(a, 1)$ divides the segment joining $(2, -5)$ and $(-4, 4)$ internally in the ratio $2 : 1$. Find $a$.
