Chapter 3.4, Problem 40PPS

To determine

To write in point-slope form for the line perpendicular to $y=-5x-8$ and contain $(-2,2)$

Answer to Problem 40PPS

Slope intercept form

  $y=\frac{1}{5}x+\frac{12}{5}$

Explanation of Solution

Given information:

Line perpendicular to $y=-5x-8$ and contain $(-2,2)$

Formula used:

Point-slope form

  $y-y_1=m(x-x_1)$

Calculation:

Let required line is $y=m_{1}x+c$

Here

  $m_1$ is slope of required line

  $c$ is y-intercept of required line

Given line

  $y=-5x-8$

  $m_2=-5$

When lines are perpendicular

Then

  $m_1\times m_2=-1$

Then slope of required line is

  $m_1\times -5=-1$

On solving

  $m_1=\frac{1}{5}$

Then slope of required line and point containing

  $m_1=\frac{1}{5}$

  $(-2,2)$

Here

  $x_1=-2$

  $y_1=2$

Using point slope formula

  $y-2=\frac{1}{5}(x-(-2))$

On solving

  $y-2=\frac{1}{5}(x+2)$

On solving

  $y-2=\frac{1}{5}x+\frac{2}{5}$

Add 2 both sides

  $y-2+2=\frac{1}{5}x+\frac{2}{5}+2$

On solving

  $y=\frac{1}{5}x+\frac{12}{5}$

Hence, slope intercept form is $y=\frac{1}{5}x+\frac{12}{5}$