Equation of a Perpendicular Bisector

A perpendicular bisector combines two straight-line skills: the midpoint (it passes through the middle) and the perpendicular gradient (it meets the segment at a right angle). Get both, then write one line.

Find the equation of the perpendicular bisector of the line segment joining $A(1, 2)$ and $B(5, 8)$, giving your answer in the form $ax + by + c = 0$. [5]

The working

Step 1, midpoint of $AB$: $M = \left(\frac{1 + 5}{2}, \ \frac{2 + 8}{2}\right) = (3, 5) \quad \text{(M1)}$

Step 2, gradient of $AB$: $m_{AB} = \frac{8 - 2}{5 - 1} = \frac{6}{4} = \frac{3}{2} \quad \text{(M1)}$

Step 3, perpendicular gradient (negative reciprocal): $m_\perp = -\frac{2}{3} \quad \text{(A1)}$

Step 4, line through $M(3, 5)$ with gradient $-\frac23$: $y - 5 = -\frac{2}{3}(x - 3)$

Clear the fraction and rearrange: $3(y - 5) = -2(x - 3) \;\Rightarrow\; 3y - 15 = -2x + 6 \;\Rightarrow\; 2x + 3y - 21 = 0 \quad \text{(M1, A1)}$

Where the marks are won and lost

• The bisector passes through the midpoint, not through $A$ or $B$. Using an endpoint gives a parallel-but-wrong line.
• The gradient is the negative reciprocal of $\frac32$, i.e. $-\frac23$, not $-\frac32$.
• The requested form $ax + by + c = 0$ needs the fraction cleared, multiply through by $3$.

Common mistakes

• Writing the line through $A(1,2)$ instead of the midpoint.
• Using $-\frac32$ (reciprocal-but-not-flipped, or sign only) for the perpendicular gradient.
• Leaving the answer as $y = -\frac23 x + 7$ when a specific form was demanded.

Full method: Perpendicular Bisector notes. Topic home: Straight-Line Graphs pillar.