Midpoints and perpendicular bisectors

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Finding the midpoint

You can find the midpoint of a line by averaging the x and y coordinates of the endpoints.

$$\begin{array}{r}A=(x_1,y_1)\\ B=(x_2,y_2)\\ \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\end{array}$$

For example we can find the midpoint of $A(1,2)$ and $B(3,4)$:

$$\begin{array}{r}A=(1,2)\\ B=(3,4)\\ \left(\frac{1+3}{2},\frac{2+4}{2}\right)=(2,3)\end{array}$$

Midpoint of a circle diameter

The midpoint of a circle’s diameter is the centre of the circle. If you have two endpoints of the diameter, you can use the same formula as above to find the centre. In this example were going to show a simple sketch:

In the segment drawn on the circle, the endpoints are $A(-4,8)$ and $B(4,4)$. The midpoint is the centre of the circle, which is $(1,6)$. The workings for this would look like this:

$$\begin{array}{r}A=(-4,8)\\ B=(4,4)\\ \left(\frac{-4+4}{2},\frac{8+4}{2}\right)=(0,6)\end{array}$$

Finding a endpoint with a midpoint

For this section were going to associate cords Q with a and b.

$$\begin{array}{r}A=(a,b)\\ M=(1,4)\\ B=(6,8)\\ \left(\frac{6+a}{2},\frac{8+b}{2}\right)=(1,4)\\ \left(\frac{6+a}{2}=1\right)\wedge \left(\frac{8+b}{2}=4\right)\\ \left(6+a=2\right)\wedge \left(8+b=8\right)\\ \left(a=2-6\right)\wedge \left(b=8-8\right)\\ A=(-4,0)\end{array}$$