Chapter 8.1, Problem 33E

Solution

The equation of the perpendicular bisector of the line segment joining the given two points.

  $y=-\frac{1}{4}x+\frac{5}{2}$

Given information:

The points given are $\left(-3,-6\right)\text{ and }\left(-1,2\right)$ .

Formula used:

1. The slope m of a line passing through the points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ , is given by: $m=\frac{y{2}_{}-y{1}_{}}{x_2-x_1}$ and slope of the perpendicular bisector:
2. The midpoint of a line passing through the points is given by: $\left(x,y\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$M
3. The equation of perpendicular bisector of the line joining two point and slope m is where c can be find by putting the middle point in this equation.

Calculation:

To calculate the equation of perpendicular bisector of the line joining two points $\left(-3,-6\right)\text{ and }\left(-1,2\right)$

First find the middle point of these two points using the formula

  

As,

  $\begin{array}{c}\left(x,y\right)=\left(\frac{-3+\left(-1\right)}{2},\frac{-6+2}{2}\right)\\ =\left(\frac{-3-1}{2},\frac{-4}{2}\right)\\ =\left(\frac{-4}{2},-2\right)\\ =\left(-2,-2\right)\end{array}$

Now find the slope of the line passing through these points

  $\begin{array}{c}m=\frac{y{2}_{}-y{1}_{}}{x_2-x_1}\\ =\frac{2-\left(-6\right)}{-1-\left(-3\right)}\\ =\frac{2+6}{-1+3}\\ =\frac{8}{2}\\ =4\end{array}$

Thus ,the slope of the perpendicular bisector is

  $-\frac{1}{m}=-\frac{1}{4}$

Thus the equation of perpendicular bisector is

  $\begin{array}{c}y=\frac{-1}{m}x+c\\ =-\frac{1}{4}x+c\end{array}$

Now find c by putting middle point in above equation

  $\begin{array}{c}y=-\frac{1}{4}x+c\\ -2=-\frac{1}{4}\left(-2\right)+c\\ -2=\frac{2}{4}+c\\ c=\frac{2}{4}+2\\ c=\frac{2+8}{4}\\ c=\frac{10}{4}\\ c=\frac{5}{2}\end{array}$

Putting value of $c=\frac{5}{2}$

  $y=-\frac{1}{4}x+\frac{5}{2}$

Thus the required equation of perpendicular bisector is

  $y=-\frac{1}{4}x+\frac{5}{2}$