Chapter 13.2, Problem 19PSB

To determine

To write: The equation of line perpendicular to the bisector of $\overleftrightarrow{\text{PR}}$ in point slope form.

Answer to Problem 19PSB

The equation of line perpendicular to the bisector of $\overleftrightarrow{\text{PR}}$ in point slope form is $y=-\frac{1}{2}x+\frac{15}{2}$ .

Explanation of Solution

Given:

The points are $P=\left(-2,5\right)$ and $R=\left(0,9\right)$ .

Calculation:

Point-slope formula is given by $m=\frac{y_2-y_1}{x_2-x_1}$ .

Find the slope of line $\overleftrightarrow{\text{PR}}$ .

Substitute $-2$ for $x_1$ , 0 for $x_2$ , $9$ for $y_2$ and $5$ for $y_1$ in the equation $m=\frac{y_2-y_1}{x_2-x_1}$ .

  $\begin{array}{c}m=\frac{9-5}{0+2}\\ =\frac{4}{2}\\ =2\end{array}$

It is given that line is perpendicular so required slope will be reciprocal of 2 that is $-\frac{1}{2}$ .

Find the mid-point of the line $\overleftrightarrow{\text{PR}}$ .

  $\begin{array}{c}\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{-2+0}{2},\frac{5+9}{2}\right)\\ =\left(-1,7\right)\end{array}$

Determine the equation of line perpendicular to the bisector of $\overleftrightarrow{\text{PR}}$ in point slope form.

Substitute $-1$ for $x_1$ , $7$ for $y_1$ , $-\frac{1}{2}$ for $m$ in the equation $y-y_1=m\left(x-x_1\right)$ .

  $\begin{array}{c}y-7=-\frac{1}{2}\left(x-1\right)\\ y-7=-\frac{1}{2}x+\frac{1}{2}\\ y=-\frac{1}{2}x+\frac{15}{2}\end{array}$

Therefore, the equation of line perpendicular to the bisector of $\overleftrightarrow{\text{PR}}$ in point slope form is $y=-\frac{1}{2}x+\frac{15}{2}$ .