Three formulas cover most basic coordinate geometry: the gradient, the midpoint and the distance. Hard questions use two or three of them in a chain, so the skill is choosing the order.
This lesson is part of SPM Additional Mathematics coordinate geometry. When you are ready for line equations, continue with parallel and perpendicular lines.
The three formulas side by side
For points A(x₁, y₁) and B(x₂, y₂):
| Quantity | Formula | Operation on coordinates |
|---|---|---|
| Gradient | (y₂ − y₁) ÷ (x₂ − x₁) | Subtract, then divide |
| Midpoint | ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2) | Add, then halve |
| Distance | √((x₂ − x₁)² + (y₂ − y₁)²) | Subtract, square, add, root |
The midpoint adds, while the gradient and distance subtract.
Worked example: a perpendicular bisector
The points are A(−1, 1) and B(5, 5). Find the equation of the perpendicular bisector of AB, then find the point C on this bisector that lies on the y-axis.
Step 1: midpoint. M = ((−1 + 5) ÷ 2, (1 + 5) ÷ 2) = (2, 3).
Step 2: gradient of AB. (5 − 1) ÷ (5 + 1) = 4 ÷ 6 = 2/3.
Step 3: perpendicular gradient. Flip and change sign: −3/2.
Step 4: equation. y − 3 = −(3/2)(x − 2). Multiply by 2: 2y − 6 = −3x + 6, so 3x + 2y = 12.
Step 5: the y-axis point. Put x = 0: 2y = 12, so y = 6 and C = (0, 6).
Check with distance. CA² = (0 + 1)² + (6 − 1)² = 1 + 25 = 26. CB² = (0 − 5)² + (6 − 5)² = 25 + 1 = 26. The two are equal, which is what a point on the perpendicular bisector must do.
The mistake that costs marks
The most common slip is using the gradient formula’s subtraction on the midpoint, or the reverse.
| Step | Wrong | Right |
|---|---|---|
| Midpoint of A(−1, 1), B(5, 5) | ((5 + 1) ÷ 2, (5 − 1) ÷ 2) = (3, 2) | ((−1 + 5) ÷ 2, (1 + 5) ÷ 2) = (2, 3) |
| Perpendicular gradient | 3/2 | −3/2 |
A quick check catches both. The midpoint must lie between the two points, and the two gradients must multiply to −1: (2/3)(−3/2) = −1.
Check yourself
The points are P(2, −1) and Q(8, 7). Find the midpoint M, the length PQ, and the equation of the line through M perpendicular to PQ.
Answer
M = ((2 + 8) ÷ 2, (−1 + 7) ÷ 2) = (5, 3).
PQ = √(6² + 8²) = √100 = 10.
Gradient of PQ = 8 ÷ 6 = 4/3, so the perpendicular gradient is −3/4.
y − 3 = −(3/4)(x − 5). Multiply by 4: 4y − 12 = −3x + 15, so 3x + 4y = 27. Check: 3(5) + 4(3) = 27.
What to study next
Continue with finding equations of parallel and perpendicular lines, then try the coordinate geometry practice set.
If you want a teacher to watch you chain these formulas on unseen questions, see online one-to-one Additional Mathematics tuition.