Midpoints and perpendicular bisectors

In a nutshell

The midpoint of a line segment can be found by finding the average of the $x$ and $y$ coordinates of its endpoints. A line segment is a section of a straight line with two distinct endpoints. A line segment joining two points on a curve is a chord. The perpendicular bisector of a chord is perpendicular to the chord and passes through its midpoint. The perpendicular bisector of a chord passes through the centre of the circle.

Midpoints

The midpoint of a line segment with endpoints ${(x_{1},y_{1})}$ and ${( x_{2}, y_{2})}$ is given by

$\boxed{\bigg(\dfrac {(x_{1} + x_{2})}{2} , \dfrac {(y_{1} + y_{2})}{2}\bigg)}$

Example 1

The chord AB of a circle has coordinates A $(4,3)$ and B $(10,3)$ . Find the coordinates of the midpoint MP of the chord AB.

In coordinate geometry questions it is helpful to draw a sketch using all the information given in the question.

Maths; Circles; KS5 Year 12; Midpoints and perpendicular bisectors

Here, ${(x_{1},y_{1})} = (4,3)$ and ${( x_{2}, y_{2})} = (10,3)$

$\bigg(\dfrac {(4 + 10)}{2} , \dfrac {(3 + 3)}{2}\bigg) = (7,3)$

Therefore, the midpoint of AB is $\underline{(7,3)}$

Perpendicular bisectors

As the perpendicular bisector of a chord is perpendicular to the chord, the product of the gradient of the chord and the perpendicular bisector is $-1$ . If the gradient of the chord is $m$ , the gradient of the perpendicular bisector is $-\dfrac {1}{m}$ .

Example 2

The chord AB of a circle has coordinates A $(3,0)$ and B $(8,-4)$ . The line $l$ is the perpendicular bisector of the chord AB, find the equation of line $l$ .

Draw a sketch using all the information given in the question.

Maths; Circles; KS5 Year 12; Midpoints and perpendicular bisectors

Here, ${(x_{1},y_{1})} = (3,0)$ and ${( x_{2}, y_{2})} = (8,-4)$ .

To find the gradient of line l, first find the gradient of AB.

$m=\dfrac {0--4}{3-8} = -\dfrac {4}{5}$

As the product of perpendicular gradients is -1 , the gradient of $l$ is $\dfrac {5}{4}$ .

To find the equation of line $l$ , the midpoint of AB can be found to obtain a set of coordinates on line $l$ .

Find the midpoint of AB.

$\bigg(\dfrac {(3 + 8)}{2} , \dfrac {(0 - 4)}{2}\bigg) = \bigg(\dfrac {11}{2},-2\bigg)$

Find the equation of line $l$ .

Use the equation ${y - y_{1} = } m (x-x_{1})$ , where ${(x_{1},y_{1})} = \bigg(\dfrac {11}{2},-2\bigg)$ ${y - y_{1} = } m (x-x_{1}$

${y - -2 = } \dfrac {5}{4} \bigg(x-\dfrac {11}{2}\bigg)$

${y +2 = } \dfrac {5}{4} x-\dfrac {55}{8}$

$\underline{{y = } \dfrac {5}{4} x-\dfrac {71}{8}}$