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. 

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

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. 

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

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

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

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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}

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

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

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