iGCSE (2021 Edition)

2.06 Perpendicular bisectors

Lesson

Perpendicular bisector

The perpendicular bisector of the line segment $AB$AB is perpendicular to $AB$AB and passes through the midpoint of $AB$AB as shown in the figure:

So to find the equation of the perpendicular bisector of the segment $AB$AB, we need the coordinates of the midpoint of $AB$AB  and the gradient of $AB$AB so that we can use the following property to find the gradient of the perpendicular of the segment:

 $m_1\times m_2$m1​×m2​ $=$= $-1$−1

Remember!

To find the equation of the perpendicular bisector of the line segment $AB$AB

• Find the midpoint of $AB$AB.
• Find the gradient of $AB$AB.
• Use the point-gradient formula for the equation of a line, $y-y_1=m(x-x_1)$yy1=m(xx1), to find the equation of the perpendicular bisector.

Worked example 1

Line segment

Find the equation of the perpendicular bisector of the line segment shown:

We can see on the diagram that the midpoint of the line segment is $(-1,1)$(1,1). And we can see that the gradient of the line segment is:

 $m$m $=$= $\frac{\text{rise}}{\text{run}}$riserun​ $=$= $-\frac{4}{6}$−46​ $=$= $-\frac{2}{3}$−23​

So we can now find the perpendicular gradient using the following method:

 $m_1\times m_2$m1​×m2​ $=$= $-1$−1 $-\frac{2}{3}\times m_2$−23​×m2​ $=$= $-1$−1 $m_2$m2​ $=$= $\frac{3}{2}$32​

Now we can substitute this gradient and the midpoint into the point-gradient formula:

 $y-y_1$y−y1​ $=$= $m(x-x_1)$m(x−x1​) $y-1$y−1 $=$= $\frac{3}{2}(x-(-1))$32​(x−(−1)) $y$y $=$= $\frac{3x}{2}+\frac{5}{2}$3x2​+52​

So the equation of the perpendicular bisector is $y=\frac{3x}{2}+\frac{5}{2}$y=3x2+52.

Worked example 2

Find the equation of the perpendicular bisector of the line segment $AB$AB where $A(-4,7)$A(4,7)  and $B(6,-3)$B(6,3).

First we have to find the midpoint of $A$A and $B$B

 Midpoint $=$= $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$(x1​+x2​2​,y1​+y2​2​) $=$= $\left(\frac{-4+6}{2},\frac{7+(-3)}{2}\right)$(−4+62​,7+(−3)2​) $=$= $\left(1,2\right)$(1,2)

Then we have to find the gradient of segment $AB$AB:

 $m$m $=$= $\frac{y_2-y_1}{x_2-x_1}$y2​−y1​x2​−x1​​ $=$= $\frac{-3-7}{6-(-4)}$−3−76−(−4)​ $=$= $\frac{-10}{10}$−1010​ $=$= $-1$−1

Therefore the gradient of the perpendicular bisector is $1$1.

So now we can use the point-gradient formula to find the equation using point $A(-4,7)$A(4,7)  and the gradient $1$1:

 $y-y_1$y−y1​ $=$= $m(x-x_1)$m(x−x1​) $y-7$y−7 $=$= $1(x-(-4))$1(x−(−4)) $y-7$y−7 $=$= $x+4$x+4 $y$y $=$= $x+11$x+11

So the perpendicular bisector is $y=x+1$y=x+1.

Outcomes

0606C8.4B

Find the equation of perpendicular bisectors.