13.01 Equations of perpendicular bisectors

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. Then we can use the following property to find the gradient of the perpendicular to the segment: 

$m_1\times m_2$m1​×m2​

$=$=

$-1$−1

 

 

Worked examples

Example 1

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

Think: We can see on the diagram that the midpoint of the line segment is $(-1,1)$(−1,1). Alternatively we could calculate the midpoint using the midpoint formula. We can also 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​

 

 

Do: 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​

 

Reflect:The gradient of a line and the gradient of the line perpendicular are called negative reciprocals. 

Do: 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))
$2y-2$2y−2	$=$=	$3x+3$3x+3
$3x-2y+5$3x−2y+5	$=$=	$0$0

 

So the equation of the perpendicular bisector is $3x-2y+5=0$3x−2y+5=0.

 

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).

Think: We need the midpoint of $AB$AB and also the gradient of the segment perpendicular to $AB$AB. 

Do: 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 midpoint $(1,2)$(1,2)  and the gradient $1$1:

$y-y_1$y−y1​	$=$=	$m(x-x_1)$m(x−x1​)
$y-2$y−2	$=$=	$1(x-1)$1(x−1)
$y-2$y−2	$=$=	$x-1$x−1
$y$y	$=$=	$x+1$x+1

 

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