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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.
  • Find the perpendicular gradient.
  • 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: 
 

$y-y_1$yy1 $=$= $m(x-x_1)$m(xx1)
$y-1$y1 $=$= $\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+x22,y1+y22)
  $=$= $\left(\frac{-4+6}{2},\frac{7+(-3)}{2}\right)$(4+62,7+(3)2)
  $=$= $\left(1,2\right)$(1,2)

 

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

$m$m $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
  $=$= $\frac{-3-7}{6-(-4)}$376(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:

$y-y_1$yy1 $=$= $m(x-x_1)$m(xx1)
$y-7$y7 $=$= $1(x-(-4))$1(x(4))
$y-7$y7 $=$= $x+4$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.

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