Perpendicular Bisectors | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Midpoint gives position; gradient gives direction

Find the midpoint of the segment. This is a point on the bisector.
Find the segment gradient.
Find the perpendicular gradient. If the segment is horizontal, the bisector is vertical; if the segment is vertical, the bisector is horizontal.
Use the midpoint in the equation of the required line.

A midpoint alone is not enough, and a perpendicular gradient alone is not enough. The first fixes where the line passes; the second fixes its direction.

A complete three-mark route

Worked example

$A(5,2)$ and $B(10,-1)$ . Find the perpendicular bisector of $AB$

(9709/12 O/N 2022 Q1(a))

$M=\left(\frac{5+10}{2},\frac{2-1}{2}\right)=\left(\frac{15}{2},\frac12\right)$
$m_{AB}=\dfrac{-1-2}{10-5}=-\dfrac35$ , so the perpendicular gradient is $5/3$ .
$y-\frac12=\frac53\left(x-\frac{15}{2}\right)$
Multiply by 6 and rearrange: $6y-3=10x-75$ , hence $5x-3y=36$ .

Answer

5x-3y=36

Swipe left or right to see the whole diagram →

The bisector passes through M(15/2, 1/2) and is perpendicular to AB.

Why every point on it is equidistant

The perpendicular bisector is also the set of all points that are the same distance from $A$ and $B$ . This is a . The two halves of the segment are mirror images across the bisector, so a point on the bisector has equal squared distances to the two endpoints.

A joins two points on a circle. The centre is equidistant from its ends, so it lies on the chord's perpendicular bisector.

Build both parts

Your turnIndependent check

Find the perpendicular bisector of the segment joining $A(-2,5)$ to $B(6,1)$ .

Show worked answer

Answer

Midpoint $M=(2,3)$ ; segment gradient $-1/2$ ; perpendicular gradient $2$ .

$y-3=2(x-2)$ , so $y=2x-1$ .

Midpoint gives position; gradient gives direction

Find the midpoint of the segment. This is a point on the bisector.
Find the segment gradient.
Find the perpendicular gradient. If the segment is horizontal, the bisector is vertical; if the segment is vertical, the bisector is horizontal.
Use the midpoint in the equation of the required line.

A midpoint alone is not enough, and a perpendicular gradient alone is not enough. The first fixes where the line passes; the second fixes its direction.

A complete three-mark route

Worked example

$A(5,2)$ and $B(10,-1)$ . Find the perpendicular bisector of $AB$

(9709/12 O/N 2022 Q1(a))

$M=\left(\frac{5+10}{2},\frac{2-1}{2}\right)=\left(\frac{15}{2},\frac12\right)$
$m_{AB}=\dfrac{-1-2}{10-5}=-\dfrac35$ , so the perpendicular gradient is $5/3$ .
$y-\frac12=\frac53\left(x-\frac{15}{2}\right)$
Multiply by 6 and rearrange: $6y-3=10x-75$ , hence $5x-3y=36$ .

Answer

5x-3y=36

Swipe left or right to see the whole diagram →

The bisector passes through M(15/2, 1/2) and is perpendicular to AB.

Why every point on it is equidistant

A joins two points on a circle. The centre is equidistant from its ends, so it lies on the chord's perpendicular bisector.

Build both parts

Your turnIndependent check

Find the perpendicular bisector of the segment joining $A(-2,5)$ to $B(6,1)$ .

Show worked answer

Answer

Midpoint $M=(2,3)$ ; segment gradient $-1/2$ ; perpendicular gradient $2$ .

$y-3=2(x-2)$ , so $y=2x-1$ .