Chapter 3.4, Problem 51PPS

To determine

To write equation of line containing $(-8,12)$ and that is perpendicular to line

containing the points $(3,2)$ and $(-7,2)$

Answer to Problem 51PPS

Slope intercept form

  $y=\frac{1}{0}(x+8)+12$

Explanation of Solution

Given information:

Line cintaining $(-8,12)$

Line perpendicular to line containing the points $(3,2)$ and $(-7,2)$

Formula used:

Slope, $m=\frac{y_2-y_1}{x_2-x_1}$

Point-slope form

  $y-y_1=m(x-x_1)$

Calculation:

Point of given line are

$(3,2)$ and $(-7,2)$

Here

  $x_1=3$

  $y_1=2$

  $x_2=-7$

  $y_2=2$

Using slope formula to calculate slope of line

  $m=\frac{2-2}{-7-3}$

On solving

  $m=\frac{0}{-10}$

On solving slope of given line

  $m=0$

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

Here

  $m_1$ is slope of required line

  $c$ is y-intercept of required line

Slope of given line

  $m_2=0$

When lines are perpendicular

Then

  $m_1\times m_2=-1$

Then slope of required line is

  $m_1\times 0=-1$

On solving

  $m_1=\frac{-1}{0}$

Then slope of required line and point containing

  $m_1=\frac{-1}{0}$

  $(-8,12)$

Here

  $x_1=-8$

  $y_1=12$

Using point slope formula

  $y-12=\frac{1}{0}(x-(-8))$

On solving

  $y-12=\frac{1}{0}(x+8)$

Add 12 both sides

  $y-12+12=\frac{1}{0}(x+8)+12$

On solving

  $y=\frac{1}{0}(x+8)+12$

Hence, slope intercept form is $y=\frac{1}{0}(x+8)+12$