Welcome to coordinate geometry — the part of the course that turns shapes into algebra by placing them on a grid. It is the single most heavily tested topic on the CSCA exam: lines, circles, and curves together make up close to a third of every paper. The good news is that almost all of it rests on a few simple tools, and this first lesson builds the most important ones.
We will start with how points sit on the plane — where they are, and how to reflect one across an axis — then learn the two formulas you will reach for again and again: the distance between two points, and the midpoint of the segment joining them. Both come straight from things you already know — Pythagoras and simple averaging — so nothing here is new from scratch. We are just putting familiar ideas onto coordinates. Get these solid and the rest of the topic becomes much easier.
You can learn anything. Let's do this!
Where a point sits
Every point on the plane is written as an ordered pair (x, y). The first number, x, says how far across the point is — right if positive, left if negative. The second, y, says how far up or down — up if positive, down if negative. The two axes cross at the origin, written O or (0, 0).
You met the four quadrants in the trigonometry topic, so we will not dwell on them. What is worth adding here is a small but slippery idea: the distance from a point to an axis.
The distance from a point to the x-axis is how far you must travel straight up or down to reach it — and that is just the size of the y-coordinate, |y|. The distance to the y-axis is how far across, |x|. This catches people out, because "distance to the x-axis" sounds like it should involve x, when in fact it is the other way round. The reason is that reaching the x-axis means changing only your height, and your height is y.
Distance to an axis. The distance from a point to the x-axis is |y| (a vertical distance). The distance from a point to the y-axis is |x| (a horizontal distance).
Worked example · finding a point from its distance (April CSCA, Q6)
A point M has x-coordinate −4, and its distance to the x-axis is 5. Find the possible coordinates of M.
The x-coordinate is handed to us: x = −4.
The distance to the x-axis is |y|, so |y| = 5.
That means y = 5 or y = −5 — five units above the axis, or five units below.
So M is (−4, 5) or (−4, −5). The distance alone cannot say which, so both are valid.
Try it! · Check-in question
A point Q has y-coordinate −3, and its distance to the y-axis is 5. What are the possible coordinates of Q?
The distance to the y-axis is the horizontal distance, |x|, so x = ±5; the y-coordinate stays fixed at −3. That gives (5, −3) or (−5, −3), so the answer is A. B swaps the roles — it lets the distance fix the y-coordinate instead of x. C and D each keep only one of the two points, but a distance cannot tell you the sign, so both ±5 are valid.
Reflecting a point
A point's reflection across an axis is its mirror image in that axis — the same distance away, but on the opposite side. Which coordinate changes depends on the mirror. Reflecting across the x-axis sends a point straight up or down to the other side, so its height flips: the y-coordinate changes sign while x stays the same. Reflecting across the y-axis is the mirror image of that — x flips, y stays. And reflecting across the origin flips both at once.
Reflecting a point. The point (a, b) reflected across the x-axis → (a, −b); across the y-axis → (−a, b); across the origin → (−a, −b). Rule of thumb: the axis you reflect across keeps its coordinate (the x-axis keeps x), and the other flips — and the origin flips both.
Worked example · reflecting across each (January 2026, Q6 · CSCA with WaWa mock, Q4)
Take the point P(3, −2). Find its reflection across the x-axis, across the y-axis, and across the origin.
Across the x-axis, the y-coordinate flips: P becomes (3, 2).
Across the y-axis, the x-coordinate flips: P becomes (−3, −2).
Across the origin, both coordinates flip: P becomes (−3, 2).
Try it! · Check-in question
The point A(−4, 5) is reflected across the x-axis. What are the coordinates of its image?
Reflecting across the x-axis flips the y-coordinate and keeps x, so (−4, 5) becomes (−4, −5). The answer is D. A (4, 5) is the reflection across the y-axis (x flips). B (4, −5) is the reflection across the origin (both flip). C (−5, −4) just swaps the two coordinates, which is not a reflection.
The distance between two points
Suppose you want the straight-line distance between two points A and B. Go across from one and up from the other, and you build a right-angled triangle whose hypotenuse is exactly the distance you are after. The horizontal leg is the gap in the x-coordinates; the vertical leg is the gap in the y-coordinates. Pythagoras finishes the job.
The distance d is the hypotenuse of a right triangle whose legs are the horizontal gap |x₂ − x₁| and the vertical gap |y₂ − y₁|. By Pythagoras, d² = (x₂ − x₁)² + (y₂ − y₁)².
If A is (x₁, y₁) and B is (x₂, y₂), the horizontal leg has length x₂ − x₁ and the vertical leg has length y₂ − y₁. By Pythagoras the distance d satisfies d² = (x₂ − x₁)² + (y₂ − y₁)², so:
Distance formula. The distance between (x₁, y₁) and (x₂, y₂) is d = √[(x₂ − x₁)² + (y₂ − y₁)²].
You never need to worry about which point you subtract first, or about negative coordinates making a gap come out negative. Each gap is squared, and squaring throws the sign away: (3 − 7)² and (7 − 3)² are both 16. Subtract whichever way is easier for you.
Worked example · distance between two points (October 2025, Q31)
Find the distance between P(−1, 2) and Q(3, 1).
The horizontal gap is 3 − (−1) = 4.
The vertical gap is 1 − 2 = −1.
Square and add: 4² + (−1)² = 16 + 1 = 17.
So d = √17. (This same pair turns up across several papers — it is a distance the exam likes.)
Try it! · Check-in question
Find the distance between A(2, −1) and B(5, 3).
The gaps are 5 − 2 = 3 across and 3 − (−1) = 4 up. Square and add: 9 + 16 = 25, and √25 = 5, so the answer is B — a 3-4-5 triangle. C (7) adds the two gaps (3 + 4) instead of using Pythagoras. A (√7) leaves 16 − 9 under the root (subtracting the squares). D (25) is the sum of squares with the square root forgotten.
Worked example · a distance that simplifies (different case)
Find the distance between A(−2, 4) and B(3, −1).
The horizontal gap is 3 − (−2) = 5.
The vertical gap is −1 − 4 = −5.
Square and add: 5² + (−5)² = 25 + 25 = 50. (The negative gap squared to a positive 25 — the sign did not matter.)
So d = √50, which tidies up: √50 = √(25 · 2) = 5√2.
Try it! · Check-in question
Find the distance between A(1, −2) and B(4, 4).
Gaps: 4 − 1 = 3 across, 4 − (−2) = 6 up. Squares 9 + 36 = 45, and √45 = √(9 · 5) = 3√5, so the answer is D. A (9) adds the gaps (3 + 6). B (3√3 = √27) subtracts the squares (36 − 9). C (5√3) swaps the numbers inside and outside the root when simplifying: 3√5 is √45, but 5√3 is √75.
Working backwards: finding a missing coordinate
Sometimes you are given the distance and one whole point, and asked for a coordinate of the other. The tool is the same distance formula — you just solve for the unknown.
And you should expect two answers, because the formula squares the unknown. This is not a quirk of the algebra; it is geometry. Every point a fixed distance from A lies on a circle around A, and a circle usually crosses a line at two points — so two positions genuinely fit.
Every point a distance d from A lies on a circle around A. That circle meets a line in two points, B₁ and B₂ — which is why solving for a coordinate gives two answers.
Worked example · solving for a coordinate (April CSCA, Q18)
The distance from A(0, 3) to B(x, −1) is 5. Find x.
Write the distance and square both sides to clear the root: (x − 0)² + (−1 − 3)² = 5².
That is x² + (−4)² = 25, so x² + 16 = 25.
Subtract 16: x² = 9.
So x = 3 or x = −3.
Two answers, because B can sit on either side of the y-axis and still be 5 away from A.
Try it! · Check-in question
The distance between A(−2, −1) and B(a, 3) is 5. What is a? (March 2026, Q17)
The vertical gap is 3 − (−1) = 4, so (a + 2)² + 16 = 25, giving (a + 2)² = 9 and a + 2 = ±3. That gives a = 1 or a = −5, so the answer is C. D (−1 or 5) solves (a − 2)² = 9 — a sign slip on the +2 inside, from writing a − (−2) as a − 2. B (±1) finds a = 1 and then assumes the other answer is −1. A (±5) finds a = −5 and assumes the other is +5. The two answers are not ± of each other: they sit 3 either side of −2.
The midpoint of a segment
The midpoint of a segment is the point exactly halfway between its two ends. Halfway across is the average of the two x-coordinates; halfway up is the average of the two y-coordinates. That is all the midpoint is — the average of the endpoints, taken one coordinate at a time.
The midpoint M lies exactly halfway along AB. Its x-coordinate is the average of x₁ and x₂; its y-coordinate is the average of y₁ and y₂.
Midpoint formula. The midpoint of the segment joining (x₁, y₁) and (x₂, y₂) is M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 ). Add the x's and halve; add the y's and halve.
Add, don't subtract. The midpoint adds the coordinates and halves; the distance subtracts them. Swapping the two is the most common slip here. If you catch yourself subtracting for a midpoint, stop — averaging means adding.
Worked example · finding a midpoint
Find the midpoint of the segment joining A(−3, 2) and B(5, 6).
Average the x-coordinates: (−3 + 5)/2 = 2/2 = 1.
Average the y-coordinates: (2 + 6)/2 = 8/2 = 4.
So the midpoint is (1, 4).
Try it! · Check-in question
Find the midpoint of the segment joining A(2, −4) and B(8, 2).
Average each coordinate: x = (2 + 8)/2 = 5, y = (−4 + 2)/2 = −1, giving (5, −1), so the answer is A. B (10, −2) forgot to halve. D (5, 1) slipped the sign when averaging −4 and 2. C (3, 3) subtracted the coordinates — that is the distance setup, not the midpoint.
Putting the two together
Many exam questions chain these tools: find a midpoint, then a distance. The classic case is the median of a triangle — the segment from a vertex to the midpoint of the opposite side. To get its length, find that midpoint first, then measure the distance to it.
Worked example · length of a median (Mock Test One, Q12)
Triangle ABC has vertices A(3, 0), B(−1, 2), and C(1, −3). Find the length of the median from A to side BC.
First, find the midpoint of BC, since that is where the median from A lands.
D = ( (−1 + 1)/2 , (2 + (−3))/2 ) = (0, −1/2).
Now measure the distance AD, from A(3, 0) to D(0, −1/2).
AD = √[(3 − 0)² + (0 − (−1/2))²] = √[9 + 1/4].
Tidy the inside: 9 + 1/4 = 36/4 + 1/4 = 37/4.
So AD = √(37/4) = √37 / 2.
Try it! · Check-in question
M is the midpoint of A(1, 1) and B(7, 9). What is the distance from the origin O to M?
First the midpoint of (1, 1) and (7, 9): M = (4, 5). Then the distance from O(0, 0) to M: √[4² + 5²] = √[16 + 25] = √41, so the answer is B. A (√130) skips the midpoint and uses B(7, 9) directly: √[49 + 81]. C (41) forgets the square root. D (9) just adds M's coordinates (4 + 5).
Your turn! · Practice questions
Practice · Question 1
Find the distance between A(−3, 1) and B(1, 5).
Gaps: 1 − (−3) = 4 across, 5 − 1 = 4 up. Squares 16 + 16 = 32, and √32 = √(16 · 2) = 4√2, so the answer is C. A (8) adds the gaps (4 + 4). B (4) uses just one gap. D (2√2 = √8) forgets to square the gaps: √(4 + 4) = √8.
Practice · Question 2
A point P lies in the second quadrant. Its distance from the x-axis is 3 and its distance from the y-axis is 4. What are its coordinates?
In the second quadrant x is negative and y is positive. The distance from the x-axis is the vertical distance |y| = 3, so y = 3; the distance from the y-axis is the horizontal distance |x| = 4, so x = −4. That gives (−4, 3), so the answer is A. B swaps which distance fixes which coordinate. C and D put the wrong sign on a coordinate for Q2.
Practice · Question 3
The distance between P(2, k) and Q(2, 7) is 5. What is k?
The x-coordinates are equal, so the distance is purely vertical: |7 − k| = 5. That gives 7 − k = 5 (so k = 2) or 7 − k = −5 (so k = 12), making the answer D. A and B each keep only one of the two solutions. C (±5) is the distance itself, not the value of k.
Practice · Question 4
M(3, −2) is the midpoint of segment AB. If A = (−1, 1), find B.
The midpoint is the average of the endpoints, so working backwards B = 2M − A. That is (2·3 − (−1), 2·(−2) − 1) = (7, −5), so the answer is A. Check: the midpoint of (−1, 1) and (7, −5) is (3, −2). ✓ B (1, −1/2) averages M and A again instead of working backwards. C (2, −1) adds M and A but forgets to double M. D (5, −3) adds A instead of subtracting it: 2M + A = (6 − 1, −4 + 1).
Practice · Question 5
Triangle ABC has A(−1, 2) and B(3, 4). The median from C to side AB has length 5, and C lies on the positive x-axis. What are the coordinates of C?
The median from C goes to the midpoint of AB: M = ( (−1 + 3)/2 , (2 + 4)/2 ) = (1, 3). Call C = (c, 0). Then CM = 5 gives (c − 1)² + (0 − 3)² = 25, so (c − 1)² = 16 and c − 1 = ±4: c = 5 or c = −3. C is on the positive x-axis, so C = (5, 0). Check: (5 − 1)² + 3² = 16 + 9 = 25. ✓ The answer is C. D (−3, 0) is the other root, but it is on the negative x-axis. B (6, 0) goes 5 straight across from M and forgets the vertical gap of 3. A (4, 0) drops M's x-coordinate: c² + 9 = 25 gives c = 4.
Things to remember
Concept
What to know
Reading a point
A point is (x, y): x is how far across, y is how far up. The origin is O = (0, 0).
Distance to an axis
Distance to the x-axis is |y| (vertical); distance to the y-axis is |x| (horizontal). It's the opposite coordinate to the one you'd expect.
Reflecting a point
(a, b) → across x-axis (a, −b); across y-axis (−a, b); across the origin (−a, −b). The mirror's own coordinate stays; the other flips.
Distance formula
d = √[(x₂ − x₁)² + (y₂ − y₁)²]. The order of subtraction and the signs don't matter — every gap is squared.
Solving for a coordinate
Substitute into the distance formula, square both sides, and solve. Expect two answers (±) — the point can sit on either side.
Midpoint formula
M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 ). Add the x's and halve, add the y's and halve.
Distance vs midpoint
Subtract the coordinates for distance; add them for the midpoint. Don't mix the two up.
Median of a triangle
Runs from a vertex to the midpoint of the opposite side. Find that midpoint first, then the distance to it.
