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 |
To find the equation of the perpendicular bisector of the line segment $AB$AB:
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.
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 | $=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−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.