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Middle Years

3.02 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. 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

 

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 examples

Example 1

Line segment

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: 
 

$y-y_1$yy1 $=$= $m(x-x_1)$m(xx1)
$y-1$y1 $=$= $\frac{3}{2}(x-(-1))$32(x(1))
$2y-2$2y2 $=$= $3x+3$3x+3
$3x-2y+5$3x2y+5 $=$= $0$0

 

So the equation of the perpendicular bisector is $3x-2y+5=0$3x2y+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+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 midpoint $(1,2)$(1,2)  and the gradient $1$1:

$y-y_1$yy1 $=$= $m(x-x_1)$m(xx1)
$y-2$y2 $=$= $1(x-1)$1(x1)
$y-2$y2 $=$= $x-1$x1
$y$y $=$= $x+1$x+1

 

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

 

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