Chapter 0.3, Problem 1QQ

To determine

The correct equation that passes through the given points and

perpendicular to the given line.

Answer to Problem 1QQ

The correct equation is option $\left(D\right)y=-1+\frac{1}{2}\left(x-3\right)$ .

Explanation of Solution

Given:

The points $:\left(3,-1\right)$

The line equation $:y=1-2\left(x+4\right)$

Options are:

  $\begin{array}{c}\left(a\right)y=3-2\left(x+1\right)\\ \left(b\right)y=-1-2\left(x-3\right)\\ \left(c\right)y=3+\frac{1}{2}\left(x+1\right)\\ \left(d\right)y=-1+\frac{1}{2}\left(x-3\right)\end{array}$

Calculation:

Given that, $y=1-2\left(x+4\right)$

Expand the equation:

  $y=1-2x-8$

Write the equation in slope-intercept form:

  $y=-2x-7$

There is a slope of $-2$ on the given line. The product of the slopes of perpendicular

line is $-1$ , so the slope of the required line is $\frac{1}{2}$ .

Find the slope between the general point $\left(x,y\right)$ and a given point $\left(3,-1\right)$ :

  $m=\frac{y-\left(-1\right)}{x-3}$

Substitute the slope of the required line $m=\frac{1}{2}$ :

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

Rewrite the equation as:

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

Thus, the equation that passes through the given points and perpendicular to the given

Line is option $\left(d\right)y=-1+\frac{1}{2}\left(x-3\right)$ .