A coordinate geometry proof has three parts: say what you will show, calculate it from the coordinates, then write a sentence that concludes. The diagram may help you plan, but it is never part of the proof.
This lesson sits in geometry problems with several possible approaches. The tools come from using gradient, midpoint and distance together.
Why can the diagram not be trusted?
Examiners draw diagrams to show the layout, not the measurements. Two sides that look equal may differ, and an angle that looks like 90° may not be.
So the argument must come from the coordinates. Write “AB² = …” and “BC² = …” and compare numbers, not impressions.
Worked example 1: an isosceles triangle with a right angle
The points are A(1, 2), B(5, 4) and C(3, 8). Show that triangle ABC is isosceles and right-angled.
Claim. Two sides are equal and they meet at a right angle.
Calculation. AB² = (5 − 1)² + (4 − 2)² = 16 + 4 = 20. BC² = (3 − 5)² + (8 − 4)² = 4 + 16 = 20. AC² = (3 − 1)² + (8 − 2)² = 4 + 36 = 40.
Conclusion. AB = BC, so the triangle is isosceles. Also AB² + BC² = 20 + 20 = 40 = AC², so by the converse of Pythagoras’ theorem the angle at B is 90°.
Worked example 2: collinear points
Show that P(−1, 1), Q(2, 7) and R(4, 11) lie on one straight line.
Gradient of PQ = (7 − 1) ÷ (2 + 1) = 6 ÷ 3 = 2. Gradient of QR = (11 − 7) ÷ (4 − 2) = 4 ÷ 2 = 2.
The two gradients are equal and both segments contain Q, so P, Q and R are collinear.
Worked example 3: a parallelogram
The points are A(0, 0), B(6, 2), C(8, 6) and D(2, 4). Show that ABCD is a parallelogram.
The midpoint of AC is (4, 3). The midpoint of BD is ((6 + 2) ÷ 2, (2 + 4) ÷ 2) = (4, 3).
The diagonals share a midpoint, so they bisect each other, and a quadrilateral whose diagonals bisect each other is a parallelogram.
The mistake that costs marks
Compare a weak answer with a full one.
| Weak answer | Full answer |
|---|---|
| “From the diagram, AB looks equal to BC.” | “AB² = 20 and BC² = 20, so AB = BC.” |
| “The gradients are the same.” | “Gradient PQ = 2 and gradient QR = 2, and Q is common, so P, Q, R are collinear.” |
| Calculation only, no last line. | A closing sentence that names the property proved. |
Check yourself
The points are P(0, 0), Q(4, 2) and R(3, 4). Show that angle PQR is a right angle and find the area of triangle PQR.
Answer
Gradient of PQ = 2 ÷ 4 = 1/2. Gradient of QR = (4 − 2) ÷ (3 − 4) = 2 ÷ (−1) = −2.
The product is (1/2)(−2) = −1, so PQ is perpendicular to QR, and angle PQR = 90°.
PQ = √(16 + 4) = √20 and QR = √(1 + 4) = √5. Area = 1/2 × √20 × √5 = 1/2 × √100 = 5 square units.
What to study next
Practise choosing the right tool in the mixed geometry practice, or see how a parameter is recovered in a line touching a curve at one point.
To have a teacher read your written arguments and show you where a mark is lost, see online one-to-one Additional Mathematics tuition.