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Locating a point in space — one more axis
◆FRAME
A city map gives two numbers for any spot — how far east, how far north. That is enough to find a shop, a bus stop, a park. It is not enough to find one flat in a twelve-storey apartment block. The map gets you to the building. It says nothing about which floor.
That third number, the floor, is not a different kind of information. It measures distance the same way the first two do: how far up from the ground. Three-dimensional space just makes this explicit. A point is not two coordinates plus an afterthought — it is three coordinates $(x, y, z)$, one signed distance along each of three mutually perpendicular axes. The third axis does not ask a new kind of question. It asks the same question, how far and which direction, one more time.
Two coordinates find the building. Three find the flat.
Nothing on the third ruler is new — which is the chapter’s opening promise, made checkable. What a flat map already asks twice, space asks once more, and the plane you already work in was never a different subject.
A street address and a house number get you to the door of the block and no further. Everything above that door shares them. The chapter’s whole move is to stop treating the floor as extra information about a place and start treating it as part of the place’s name — which is what the third number in a coordinate triple is.
Ask for a flat by street and house number alone and you have named the whole staircase, not one home. That is what a two-number address does in space — and it is why the third coordinate is not an extra.
Three mutually perpendicular axes, X, Y and Z, meet at a fixed ORIGIN $O$. Each PAIR of axes marks out a COORDINATE PLANE: the XY-plane is the floor, the YZ-plane and ZX-plane stand up from it like two walls meeting at a corner.
In two dimensions, one axis crossing another cuts the plane into four quadrants. Three planes crossing each other do the same job with one more plane in play, and the arithmetic of “one more plane” is not four pieces but eight. Space splits into EIGHT OCTANTS, one for every combination of sign a point’s three coordinates can carry: $(+,+,+)$, $(+,+,-)$, and so on down to $(-,-,-)$.
A building standing at the corner of two roads already lives in this pattern. Which side of each road fixes $x$ and $y$; whether you are above or below the ground floor fixes $z$. Three yes-or-no questions, eight possible answers.
The word octant is worth pausing on. It is not a new idea with a new name: it is quadrant with one more halving. Each plane you add asks a point one more yes-or-no question — which side? — and each answer doubles the count. Four became eight for the same reason two became four when the y-axis was drawn. Nothing about a point changes when it sits in the seventh of them rather than the first.
A point $P$ in space carries three coordinates, $(x, y, z)$ — each one a signed perpendicular distance from a coordinate plane, not from an axis. $x$ is measured from the YZ-plane, $y$ from the ZX-plane, $z$ from the XY-plane.
In the FIRST octant, the direct 3D analogue of the first quadrant, all three coordinates read positive. Step across any one plane and exactly that one coordinate flips sign — the other two are untouched, because crossing a wall changes only the distance measured from that wall.
This is the flat-and-floor idea again, spelled out in full. $x$ and $y$ place the building on the plot; $z$ places the flat on the building. Change the floor and only $z$ moves.
Each coordinate answers one question, from one plane. Never mix them up.
Notice which letter measures from which plane, because the pairing is not the one a reader expects. It is easy to carry over from two dimensions the habit of thinking of x as a distance along the x-axis. Here it is a distance from the plane the x-axis is NOT in. Getting that pairing wrong is the quiet source of most wrong answers in this chapter.
Each crossing is cut in half by its own wall — the equal ticks say so — so the distance keeps its size and only swaps side. The other two coordinates are measured from the other two walls, and a step across this one never reaches them.
Worked example
Which octant holds the point $(2, -3, 4)$?
$x = 2$, $y = -3$, $z = 4$ read the three signed coordinates straight off the point
$x$ positive, $y$ negative, $z$ positive the sign of each coordinate is all that decides the octant — the size of the number never matters
not the first octant the first octant needs all three coordinates positive; $y$ fails that test here
$(2, -3, 4)$ sits below the ZX-plane a negative $y$ places the point on the far side of the ZX-plane from every first-octant point
The minus sign is a direction, not a defect. Both points are 2 along x, 3 from the zx-plane and 4 up — the same three distances. The y-step is walked backward for P and forward for P', and that single reversal is the whole difference between them. Count the positive coordinates before naming an octant: three positives means octant I, and P has only two, so it is ruled out of the first octant and out of nothing else.
To name the octant, read the three signs and stop there. Three of these points sit on one ray, so only their size differs; the fourth lies well off it. Neither the size nor the direction changes the answer.
Two flats in two different towers, at two different floors — how far apart are they, in a straight line through the air? Ground distance alone will not answer that. A drone flying from one balcony to the other also climbs or drops between floors, and that climb is part of the real distance too.
The formula carries all three differences at once. For $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the distance is:
It is the familiar two-dimensional distance formula with one more squared difference added — the same Pythagorean idea, run through three independent directions instead of two.
Drop $z$ and the third term vanishes on its own, since $z_2 - z_1$ becomes 0 when both points share a floor. The 3D formula does not replace the 2D one; it contains it.
The right angle that licenses the second use of Pythagoras is the one you cannot see. A climb straight up leaves the floor at ninety degrees to every line lying in it — including the diagonal walk you have just measured — and that is exactly the condition the theorem asks for. The box is drawn so that you do not have to take that second right angle on trust: its faces are where it lives.
The map keeps only the base of each triangle, and that alone is enough to put the two pairs in the wrong order. The climb is not a correction applied afterwards to a distance — it is part of the distance, and here it is the part that decides.
Worked example
Find the distance between $(1, 2, 3)$ and $(4, 6, 3)$
$d = \sqrt{(4-1)^2 + (6-2)^2 + (3-3)^2}$ substitute the two points straight into the distance formula, in order
$d = \sqrt{9 + 16 + 0}$ square each difference — the $z$ difference is 0, since both points share the same $z$
$d = \sqrt{25} = 5$ add and take the root — with $z$ contributing nothing, this is really a 2D distance in disguise
Distance-from-origin, next, runs this same idea one step further: not a coordinate that happens to match between two points, but a whole point that happens to be (0, 0, 0), so every one of the three differences collapses to the coordinate itself rather than to zero. The formula does not change; only what gets subtracted does.
A drone’s flight computer tracks one number constantly: how far it now sits from its own launch point, wherever the flight has taken it. That launch point is the natural origin for the whole flight, and the distance back to it is a special case of the formula just built.
Set $(x_1, y_1, z_1) = (0, 0, 0)$ in the general distance formula and every difference collapses to a single coordinate: $x_2 - 0 = x_2$, and the same for $y$ and $z$. What is left is the distance of any point $(x, y, z)$ from the origin:
$d = \sqrt{x^2 + y^2 + z^2} \cdot$
Each coordinate is squared on its own, with nothing left to subtract.
Nothing new was proved here. The general formula was simply asked a simpler question.
Notice that the origin did no work here that a different choice of origin could not have done somewhere else. The corner of a room, the base of a mast, a surveyor’s benchmark: whenever a problem is free to put O where it likes, it should put it at the point everything is being measured from, and the three subtractions stop being work. That choice is made before any arithmetic starts, and it is the only part of this the formula cannot do for you.
A new mobile tower is going up on the straight line between two existing towers, at a point that splits the distance in a chosen ratio. That point sits closer to one tower than the other, by exact design, not by guesswork.
Take $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, and divide segment $P Q$ INTERNALLY in the ratio $m:n$, strictly between $P$ and $Q$. The dividing point’s coordinates are given by the SECTION FORMULA:
$((m x_2 + n x_1)/(m+n), (m y_2 + n y_1)/(m+n), (m z_2 + n z_1)/(m+n)) \cdot$
Each coordinate is worked out the same way: a weighted average of the two endpoints’ values. The SAME weights $m$ and $n$ apply independently to $x$, $y$ and $z$ — get one coordinate right and the other two follow the identical pattern.
One ratio, applied three times over — once per coordinate, never mixed.
The reason the section formula has three lines rather than one is that nothing in it couples the coordinates together. Each line is the same weighted average done to a different number, and none of the three can be affected by what the other two are doing. That independence is why the two-dimensional formula extended without needing a new argument.
The dividing point is not a fixed place on PQ — the ratio chooses it. Run the formula three times on one segment and all three answers land on it. The midpoint is simply the case where the two parts come out equal.
Worked example
Divide the segment from $P(2, 1, -3)$ to $Q(6, 5, 9)$ internally in the ratio $1:2$
$m = 1$, $n = 2$ read the ratio in order — $m$ is the weight nearer $Q$, $n$ the weight nearer $P$
$x = (1(6) + 2(2))/3 = 10/3$ apply the section formula to the $x$-coordinates alone, dividing by $m + n = 3$
$y = (1(5) + 2(1))/3 = 7/3$ repeat for $y$ — same weights, same denominator, new coordinates
$z = (1(9) + 2(-3))/3 = 3/3 = 1$ repeat for $z$ — the negative coordinate carries its sign straight through the arithmetic
$(10/3, 7/3, 1)$ the three results together are the dividing point
The two coloured lengths are the ratio itself: one part from P, two parts to Q. You can check the division against them without reading a number, which is the point of drawing it. The red dot is where the same formula lands if the ratio is written the other way round — same segment, same endpoints, a point nowhere near the right one. The rule that decides it: m always multiplies the far endpoint, Q.
Two of this answer’s three coordinates are thirds, which is exactly where slips happen. The z coordinate is whole, so use it: compare how far R has risen above P with the whole rise from P to Q.
▲CONCEPT
Split a segment exactly in half and the ratio is $1:1$ — no tower sits closer to one end than the other. The MIDPOINT of $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is nothing but the internal section formula with $m:n = 1:1$:
$((x_1+x_2)/2, (y_1+y_2)/2, (z_1+z_2)/2) \cdot$
Each coordinate is the plain average of its two endpoints. An equal ratio weights both endpoints the same, so the weighted average the section formula always computes simplifies to an ordinary one. No new formula was needed — the general case already contained this one.
Not every marker sits between two known points. A surveyor extending a straight pipeline past its second known marker needs a THIRD point beyond it, at a stated ratio to the first two. That point sits outside the segment, on the line running through it.
Take $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. The point dividing segment $P Q$ EXTERNALLY in the ratio $m:n$ lies outside segment $P Q$, on the line through $P$ and $Q$ extended. Its coordinates are:
$((m x_2 - n x_1)/(m-n), (m y_2 - n y_1)/(m-n), (m z_2 - n z_1)/(m-n)) \cdot$
That is the same formula as internal division, with every $n$ replaced by $-n$. That single sign flip is exactly what “external” means algebraically: the second point’s pull on the answer now works in reverse.
Same formula, one sign reversed — because the point being sought has moved from between the two markers to beyond them.
This is the same picture as the last one with a single number changed, and it is worth reading them side by side. Nothing about the segment, the endpoints or the ratio has moved. Only the sign carried by the second part has, and the answer left the segment entirely. A sign is not a decoration on this formula; it is the formula’s whole geometry.
The margin’s warning that external division needs m and n to differ is not really a rule about dividing by zero. Watch where the answer goes as the two numbers close on each other, and you can see what it is protecting you from.
Worked example
Divide the segment from $P(2, 1, 3)$ to $Q(4, 3, 9)$ externally in the ratio $2:1$
$m = 2$, $n = 1$ read the ratio in order, exactly as for internal division
$x = (2(4) - 1(2))/1 = 6$ apply the external section formula to $x$ — the denominator is $m - n = 1$
$y = (2(3) - 1(1))/1 = 5$ repeat for $y$, same weights and denominator
$z = (2(9) - 1(3))/1 = 15$ repeat for $z$ — the result overshoots both original points, which is expected outside the segment
$(6, 5, 15)$ the dividing point lies beyond $Q$ on the extended line, not between $P$ and $Q$
At the ratio 2 : 1 the denominator m minus n comes out 1, and the whole formula collapses to one instruction: take the step from P to Q, then take it again. R lands past Q because you did not stop there.
Every external-division answer can be sanity-checked this cheaply, and it costs one line of arithmetic. Do the coordinate with the friendliest numbers first — if it lands between the two endpoints, you have solved the wrong problem.
Every idea in this chapter answers one question: where, exactly, in space?
Three signed coordinates locate a point against the three coordinate planes — the flat-and- floor idea from the start. The distance formula measures how far apart two located points sit, or how far one sits from the origin, by carrying the two-dimensional formula through one more independent direction. The section formulas locate a NEW point on the line through two known points, at a chosen ratio — internal when that point sits between them, external when it sits beyond.
Notice the pattern underneath all four results. Every 3D formula here is its matching 2D formula with one more coordinate carried through, unchanged in structure. Nothing about space in three dimensions demanded new mathematics. It only asked the old questions once more, along an axis that was always there to be asked.
One more axis, the same tools — asked three times instead of two.
THE TRAP. A point in three-dimensional space always has three positive coordinates.
THE REALITY. Only points in the FIRST octant have all three coordinates positive — exactly like a point in two dimensions is only guaranteed positive coordinates in the first quadrant.
A point such as $(-2, 5, -1)$ is a perfectly ordinary point in space, sitting in one of the other seven octants. Nothing about it is broken or exceptional. It fails the all-positive test the same way a point in the second quadrant fails it in two dimensions. One lies on the wrong side of an axis, the other on the wrong side of a plane — not in error, just elsewhere.
Positive is one octant’s property. It was never the whole space’s rule.
Nobody thinks (-2, 5) is an illegal point on a flat grid — it is quadrant II, and it has been allowed since Class 9. Drop three below the plane and one more sign appears; no new kind of thing does. What changed is only the requirement: quadrant I asks for two positives, octant I asks for three, and this point has one either way. A negative coordinate rules a point out of the first octant. It rules it out of nothing else.
✕MISCONCEPTION
THE TRAP. The external section formula is the internal one with the same plus signs — just use the ratio $m:n$ in a different order.
THE REALITY. External division REPLACES $n$ with $-n$ throughout the internal formula. Every $+n(\ldots)$ term becomes $-n(\ldots)$, and the denominator becomes $m - n$ instead of $m + n$.
Drop that sign change and the arithmetic still runs to completion — it simply computes an INTERNAL point again, silently, never the external one the problem actually asked for. The calculator gives no warning. Only checking whether the answer lies outside segment $P Q$, as an external point must, catches the mistake.
_Reordering the ratio changes nothing. Flipping the sign on $n$ is the whole of the rule._
Both readings of the ratio put the answer between P and Q — one a third along, the other two thirds. That is not a coincidence: a weighted average of two numbers can never leave the interval they span, so no reordering of m and n will ever produce a point off the segment. Only replacing n with minus n does that, which is why the denominator changes from m + n to m − n as well. Note too where the external point landed: beyond P, not beyond Q. External means outside, not past the far end.
External division is not internal division with m and n swapped; swapping still lands you inside. Replacing n with -n is what turns the denominator from m + n into m - n, and that is what carries the point past Q.
practice In which octant does the point $(-3, 4, -2)$ lie?
I
III
VI
VIII
Answers
VI
Exercise 11.1 hands you three numbers and asks which octant. This is why the question has an answer at all — the sizes are no help, because four different points carry exactly those sizes. Read the signs.
Exercise 11.2 — The distance formula
practice Find the distance between the points $A(2, 3, 1)$ and $B(4, 6, 7)$.
practice Show that the points $A(1, 2, 3)$, $B(-1, -1, -1)$, and $C(3, 5, 7)$ are collinear, using the distance formula.
Answers
7
$A B = A C = \sqrt{29}$, $B C = 2 \sqrt{29}$; collinear.
Exercise 11.2 asks you to decide whether three points lie on one line and hands you only a distance formula to do it with. This is why that works. If B sits on AC, the two short walks are the long walk, so their lengths add exactly. Step B off the line and A to B to C becomes a detour — and a detour is always longer, so the sum overshoots. The test is not a trick; it is the triangle inequality with the slack taken out.
Miscellaneous — The section formula
practice Find the coordinates of the point which divides the line segment joining $P(4, -3, 6)$ and $Q(-2, 9, -6)$ internally in the ratio $1:2$.
Answers
$(2, 1, 2)$
Before trusting the arithmetic, check the picture: the answer has to sit ON the segment, one third of the way from P, and nowhere else. This item is a good one to check that way because its three coordinates do different things — x falls, y rises, z falls — and a reader who gave m to the wrong endpoint lands two thirds along instead, which the eye catches at once.
A ratio here is a ratio of distances, not just a rule for weighting numbers. This item is the rare one where both distances come out whole, so you can see for yourself what 1 : 2 is measuring.
board-easy How many coordinates fix a point in three-dimensional space?
$2$
$3$
$4$
$6$
board-easy State the sign of each coordinate of $(2, -3, 4)$, and say whether it lies in the first octant.
board-easy Find the point’s distance from the XY-plane if the point is $(4, -1, -6)$.
board-easy Find the distance between the points $(2, 3, 1)$ and $(5, 7, 1)$.
board-easy Find the distance of the point $(6, 2, 3)$ from the origin.
board-easy Find the midpoint of $(2, -4, 6)$ and $(8, 2, -2)$.
board-easy Find the point dividing the segment from $(1, -1, 2)$ to $(9, 3, -2)$ internally in the ratio $1:3$.
board-easy Find the point dividing the segment from $(3, -1, 2)$ to $(5, 3, -4)$ externally in the ratio $3:1$.
board-easy Three mutually perpendicular coordinate planes cut space into how many octants?
$4$
$6$
$8$
$10$
board-easy Find the distance of $(-1, -2, -2)$ from the origin.
board-standard Points $(2, 3, 4)$, $(5, 7, 10)$ and $(8, 11, 16)$ — show they are collinear, using the distance formula.
board-standard Find the point on the x-axis that is equidistant from $(1, 2, 3)$ and $(3, 2, -1)$.
board-standard In what ratio does the point $(2, 5, -1)$ divide the segment from $(1, 3, 2)$ to $(4, 9, -7)$?
board-standard One endpoint of a segment is $(3, -2, 5)$ and its midpoint is $(6, 1, 3)$. Find the other endpoint.
board-standard Show that the triangle with vertices $(0, 0, 0)$, $(1, 2, 2)$ and $(2, -2, 1)$ is right-angled at the origin.
board-standard Find the point dividing the segment from $(1, 2, -3)$ to $(4, 5, 6)$ externally in the ratio $4:1$.
board-standard Verify that the origin is the midpoint of $(1, 1, 1)$ and $(-1, -1, -1)$.
board-standard A student claims every point in 3D space has three positive coordinates. Give one point that disproves this, and state which coordinate is negative.
board-standard A student divides the segment from $(1, 0, 2)$ to $(5, 4, -2)$ in ratio $3:1$ using the internal formula, then calls the result the external point. Find the correct external point instead.
board-standard The midpoint of $(2, -3, 4)$ and $(x, 5, 8)$ is $(5, 1, 6)$. Find $x$.
$6$
$7$
$8$
$9$
board-standard If the distance of the point $(a, 4, 3)$ from the origin is $\sqrt{50}$, find the positive value of $a$.
board-standard A drone flies in a straight line from $(0, 0, 0)$ to $(12, 9, 6)$. Find its position after covering exactly one-third of the distance.
JEE Show that the triangle $(0, 0, 0)$, $(3, 4, 0)$, $(0, 4, 3)$ is isosceles.
JEE Find the ratio in which the YZ-plane divides the segment joining $(-2, 4, 7)$ and $(3, -5, 8)$.
JEE Find the ratio in which the plane $y=0$ divides the segment joining $(1, -1, 4)$ and $(3, 5, 2)$.
JEE Three vertices of a parallelogram are $(3, -1, 2)$, $(1, 2, -4)$ and $(-1, 1, 2)$. Find the fourth vertex, opposite the second one.
JEE Find the point on the y-axis equidistant from $(3, 1, 2)$ and $(5, -3, 4)$.
JEE A point $R$ divides the segment from $(2, -3, 4)$ to $(8, 0, 10)$ such that the near-side length is $2/5$ of the whole segment. Find the x-coordinate of $R$.
$22/5$
$18/5$
$26/5$
$4$
JEE Show that $(1, 2, 3)$, $(2, 3, 1)$ and $(3, 1, 2)$ form an equilateral triangle.
JEE For $(0, 0, 0)$ and $(9, 9, 9)$, the section formula gives the point $(3, 3, 3)$ dividing the segment internally in ratio $1:2$. Confirm the two shorter distances sum to the whole, using the distance formula.
Answers
$3$
x pos, y neg, z pos; not first octant
$6$
$5$
$7$
$(5, -1, 2)$
$(3, 0, 1)$
$(6, 5, -7)$
$8$
$3$
collinear — $\sqrt{61} + \sqrt{61} = 2 \sqrt{61}$
$x=0$, the origin
$m:n = 1:2$
$(-3, 11, 10)$
right-angled — 9 + 9 = 18
$(5, 6, 9)$
$(0, 0, 0)$
$(4, -6, 3)$, y negative
$(7, 6, -4)$
$8$
$5$
$(4, 3, 2)$
isosceles — both sides equal 5
$m:n = 2:3$
$m:n = 1:5$
$(1, -2, 8)$
$y=-9/2$
$22/5$
equilateral — all sides equal $\sqrt{6}$
$3 \sqrt{3} + 6 \sqrt{3} = 9 \sqrt{3}$
This is the chapter-end item whose method is not the one it looks like. The angle at O lies in no coordinate plane, so there is nothing on the page to measure and no drawing of it can be read off. Three distances settle it instead: the two legs come out 3 and 3, the opposite side 3 root 2, and 9 + 9 = 18. That is Pythagoras run backwards — if the squares add, the angle between them is right.
This item reads like a search along the x-axis. Put the unknown point at (a, 0, 0), set the two squared distances equal, and the a-squared terms cancel, leaving 4a = 0. The point you are looking for is the origin.
2:1 — 2:1 inverts the correct ratio — check which endpoint the dividing point sits closer to.
1:3 — 1:3 does not satisfy the x-coordinate equation; resolve it directly.
3:1 — 3:1 does not satisfy the x-coordinate equation; resolve it directly.
Find the point dividing the segment from (2, 3, -1) to (6, 7, 3) externally in the ratio 3:1.
(5, 6, 2)
(8, 9, 5)
(-8, -9, -5)
(6, 7, 3)
Check your answer
(5, 6, 2) — (5, 6, 2) is the internal division point, not the external one — the sign of n must flip.
✓ (8, 9, 5) — (B) x = (3(6)-1(2))/2 = 8, y = (3(7)-1(3))/2 = 9, z = (3(3)-1(-1))/2 = 5.
(-8, -9, -5) — (-8, -9, -5) flips the denominator’s sign, using n minus m instead of m minus n.
(6, 7, 3) — (6, 7, 3) is the second given point itself, not the dividing point.
Find the ratio in which the YZ-plane divides the segment joining (-2, 4, 7) and (3, -5, 8).
3:4
3:2
2:3
1:1
Check your answer
3:4 — 3:4 does not satisfy the x-coordinate equation set to zero; recheck the algebra.
3:2 — 3:2 inverts the correct ratio — resolve the x-coordinate equation directly, without guessing the order.
✓ 2:3 — (C) The YZ-plane is x = 0, so (3m-2n)/(m+n) = 0 gives 3m = 2n, so m:n = 2:3.
1:1 — 1:1 does not satisfy the x-coordinate equation set to zero; recheck the algebra.
Find the ratio in which the plane y = 0 divides the segment joining (2, -5, 1) and (4, 3, 9).
3:8
3:5
1:1
5:3
Check your answer
3:8 — 3:8 does not satisfy the y-coordinate equation set to zero; recheck the algebra.
3:5 — 3:5 inverts the correct ratio — resolve the y-coordinate equation directly, without guessing the order.
1:1 — 1:1 does not satisfy the y-coordinate equation set to zero; recheck the algebra.
✓ 5:3 — (D) The plane y = 0 gives (3m-5n)/(m+n) = 0, so 3m = 5n, so m:n = 5:3.
Find the ratio in which the plane z = 0 divides the segment joining (1, 2, -6) and (5, -2, 4).
3:2
2:3
1:2
2:1
Check your answer
✓ 3:2 — (A) The plane z = 0 gives (4m-6n)/(m+n) = 0, so 4m = 6n, so m:n = 3:2.
2:3 — 2:3 inverts the correct ratio — resolve the z-coordinate equation directly, without guessing the order.
1:2 — 1:2 does not satisfy the z-coordinate equation set to zero; recheck the algebra.
2:1 — 2:1 does not satisfy the z-coordinate equation set to zero; recheck the algebra.
Find the point on the x-axis equidistant from (2, 3, -1) and (6, 3, 3).
(4, 0, 0)
(5, 0, 0)
(0, 5, 0)
(0, 0, 5)
Check your answer
(4, 0, 0) — (4, 0, 0) does not satisfy the equidistance equation; resolve it directly.
✓ (5, 0, 0) — (B) Setting the two squared distances equal gives 8x = 40, so x = 5, giving (5, 0, 0).
(0, 5, 0) — (0, 5, 0) puts the unknown value in the y-slot — a point on the x-axis needs it in the x-slot.
(0, 0, 5) — (0, 0, 5) puts the unknown value in the z-slot — a point on the x-axis needs it in the x-slot.
Are (1, -1, 3), (2, -4, 5) and (4, -10, 9) collinear? Check whether the distance between the first two, plus the distance between the last two, equals the distance between the first and third.
Yes, but only because two of the points coincide.
No — the three distances do not satisfy the sum condition.
Yes, collinear — the distances sum correctly.
Cannot be determined from the distance formula alone.
Check your answer
Yes, but only because two of the points coincide. — None of the three points coincide; check each pair’s coordinates directly.
No — the three distances do not satisfy the sum condition. — The three computed distances DO satisfy the sum condition — recompute each one before rejecting it.
✓ Yes, collinear — the distances sum correctly. — (C) The two shorter distances are the square root of 14 and 2 times the square root of 14, summing to 3 times the square root of 14, which equals the longest distance.
Cannot be determined from the distance formula alone. — The distance formula always settles collinearity this way — three points are collinear exactly when the two shorter distances sum to the longest.
Is the triangle with vertices (0, 0, 0), (2, 1, 2) and (2, -2, -1) right-angled? Identify the right-angle vertex.
No, it is not right-angled.
Yes, at (2, 1, 2).
Yes, at (2, -2, -1).
Yes, at the origin.
Check your answer
No, it is not right-angled. — Pythagoras IS satisfied here — 9 plus 9 equals 18, so the triangle is right-angled.
Yes, at (2, 1, 2). — The two sides meeting at (2, 1, 2) do not satisfy Pythagoras; check the origin instead.
Yes, at (2, -2, -1). — The two sides meeting at (2, -2, -1) do not satisfy Pythagoras; check the origin instead.
✓ Yes, at the origin. — (D) The two sides from the origin each have squared length 9, and the third side has squared length 18, so 9+9=18 and the right angle sits at the origin.
Triangle (0, 0, 0), (2, 2, 1), (2, -2, -1) — classify it by side lengths.
Isosceles
Equilateral
Scalene
Cannot classify without more information
Check your answer
✓ Isosceles — (A) Two sides from the origin each have squared length 9; the third side has squared length 20 — isosceles, not equilateral.
Equilateral — The third side has squared length 20, not 9 — only two sides are equal, not all three.
Scalene — Two sides here DO have the same squared length, 9 — recompute all three side lengths.
Cannot classify without more information — The three side lengths, once computed, settle the classification directly.
A(2, 3, 1), B(6, 7, 2), C(8, 3, 7) are three vertices of a parallelogram ABCD. Find D.
(4, 1, 6)
(4, -1, 6)
(12, 13, 10)
(-4, 1, -6)
Check your answer
(4, 1, 6) — (4, 1, 6) gets two coordinates right but slips the sign on the middle one.
✓ (4, -1, 6) — (B) The diagonals of a parallelogram share a midpoint — solving from the midpoint of A and C gives D = (4, -1, 6).
(12, 13, 10) — (12, 13, 10) sums all three vertices directly — the diagonals' shared midpoint step was skipped.
(-4, 1, -6) — (-4, 1, -6) negates every coordinate of the correct fourth vertex.
A relay signal travels in a straight line from tower A(0, 0, 100) to tower B(300, 400, 100) metres. Find the position of a repeater placed one-quarter of the way from A to B.
(150, 200, 100)
(225, 300, 100)
(75, 100, 100)
(100, 400/3, 100)
Check your answer
(150, 200, 100) — (150, 200, 100) is the midpoint, halfway along — not one-quarter of the way.
(225, 300, 100) — (225, 300, 100) is the point three-quarters of the way, using the reversed ratio 3:1.
✓ (75, 100, 100) — (C) One-quarter of the way uses ratio 1:3, giving x=(1(300))/4=75, y=(1(400))/4=100, z=100.
(100, 400/3, 100) — (100, 400/3, 100) divides by 3 instead of the correct total of 4.
Which formula correctly gives the distance from (x, y, z) to the origin?
the square root of ((x-1) squared plus (y-1) squared plus (z-1) squared)
x squared plus y squared plus z squared
the square root of (x plus y plus z)
the square root of (x squared plus y squared plus z squared)
Check your answer
the square root of ((x-1) squared plus (y-1) squared plus (z-1) squared) — There is no reason to subtract 1 from each coordinate for a distance from the origin.
x squared plus y squared plus z squared — x squared plus y squared plus z squared is the value under the root, not the distance itself.
the square root of (x plus y plus z) — The square root of x plus y plus z does not compute a distance at all — each coordinate must be squared first.
✓ the square root of (x squared plus y squared plus z squared) — (D) The origin-distance formula squares each coordinate, adds them, then takes one square root.
The point (4, 5, 6) divides the segment from P(1, 2, 3) to Q(3, 4, 5) externally, in ratio m:n. Find m:n.
1:3
1:2
3:1
2:1
Check your answer
1:3 — 1:3 inverts the correct ratio — check the x-coordinate equation directly rather than guessing which point is 'closer'.
1:2 — 1:2 does not satisfy the x-coordinate equation; resolve it directly.
✓ 3:1 — Setting the external formula’s x-coordinate equal to 4 gives 3m - n = 4(m-n), so m = 3n, i.e. m:n = 3:1.
2:1 — 2:1 does not satisfy the x-coordinate equation; resolve it directly.
The step that decides these items is not the algebra, it is the translation. "The yz-plane" has to become "x = 0" before there is anything to solve. Once it does, the problem is one equation: set the x-numerator to zero, and the ratio falls out — the other two coordinates never enter it. The picture is the equation: the only thing true of every point on that sheet is that its first coordinate is nothing.
All four options are real numbers you can compute from (2, 2, 1); only one of them is a distance from the origin, and drawing the four to one scale is what separates them.
The whole value of this item is noticing that nothing has to be squared. The three differences are 4, 3 and 12. Four and three fix the floor walk at 5 by a triangle you have known since Class 7; five and twelve then fix the answer at 13 by the next one along. Two memorised triples, stacked, the second standing on the first one’s answer. Squaring and adding gets 169 and the same 13 — thirty seconds later.
Knowing which property to reach for is the whole item. Diagonals that bisect each other turn a fourth-vertex question into a single equation, and that equation rearranges into a subtraction you can do without writing anything down.