Coordinate Geometry • Topic 1 of 4

Distance & Midpoint

The distance between A(x₁, y₁) and B(x₂, y₂) is just Pythagoras on the right triangle whose legs are the horizontal gap (x₂−x₁) and the vertical gap (y₂−y₁): d = √[(x₂−x₁)² + (y₂−y₁)²]. Because the differences are squared, the order of the points never matters and the answer is always non-negative. The midpoint is even simpler — average the x-coordinates and average the y-coordinates: M = ((x₁+x₂)/2, (y₁+y₂)/2). Two CAT-useful facts fall straight out of these. First, equal distances let you prove a triangle is isosceles or equilateral or find a point on an axis equidistant from two given points (set the two distance expressions equal and the square roots drop away). Second, the midpoint formula is your fast route to the circumcentre of a right triangle (the midpoint of the hypotenuse) and to checking whether a quadrilateral is a parallelogram (diagonals share a midpoint). Keep the Pythagorean triples 3-4-5, 5-12-13, 8-15-17 in mind — many "find the distance" answers land on one of them.

✅ Solved examples

1. Find the distance between A(2, 3) and B(6, 6).
Δx = 4, Δy = 3, so d = √(16 + 9) = √25 = 5. (A 3-4-5 triangle.)
2. Find the midpoint of the segment joining (−3, 7) and (5, −1).
M = ((−3+5)/2, (7+(−1))/2) = (2/2, 6/2) = (1, 3).
3. A point on the x-axis is equidistant from (1, 5) and (4, −4). Find it.
Let it be (x, 0). (x−1)²+25 = (x−4)²+16 ⇒ −2x+26 = −8x+32 ⇒ 6x = 6 ⇒ x = 1. Point is (1, 0).
4. Show that (0, 0), (3, 4) and (−4, 3) are vertices of an isosceles right triangle, and find its area.
Sides: from origin to (3,4) is 5, to (−4,3) is 5; between (3,4) and (−4,3) is √(49+1) = √50 = 5√2. Two legs of 5 with hypotenuse 5√2 ⇒ isosceles right triangle. Area = ½·5·5 = 12.5.

✏️ Practice — try these, take hints as needed

1. Find the distance between (−1, −1) and (2, 3).
Δx = 3, Δy = 4.
√(Δx² + Δy²).
√(9 + 16).
5
2. Find the midpoint of (8, −2) and (−2, 10).
Average the x-values.
Average the y-values.
((8−2)/2, (−2+10)/2).
(3, 4)
3. The point (a, 0) is equidistant from (2, −5) and (−2, 9). Find a.
Set the two squared distances equal.
(a−2)²+25 = (a+2)²+81.
Expand: −8a = 56.
a = −7
4. If (3, k) is at distance 5 from (0, 1), find k.
9 + (k−1)² = 25.
(k−1)² = 16.
k − 1 = ±4.
k = 5 or k = −3
5. A circle has diameter endpoints (−2, 3) and (4, −5). Find its centre and radius.
Centre is the midpoint.
Radius is half the diameter length.
Diameter = √(36+64) = 10.
Centre (1, −1), radius 5

📝 Topic test — 8 questions

Auto-graded with full solutions; saved to your dashboard. Use the calculator and formula sheet (top-right) any time.

Loading questions…
Section Formula →