Chapter 3.4, Problem 10CYU

To determine

To write in point-slope form for the line perpendicular to $y=-2x+6$ and contain $(3,2)$

Answer to Problem 10CYU

Slope intercept form

  $y=-\frac{1}{2}x+\frac{9}{2}$

Explanation of Solution

Given information:

Line perpendicular to $y=-2x+6$ and contain $(3,2)$

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

  $y=\text{ -2}x+6$

  $m_2=-2$

When lines are perpendicular

Then

  $m_1\times m_2=-1$

Then slope of required line is

  $m_1\times -2=-1$

On solving

  $m_1=\frac{1}{2}$

Then slope of required line and point containing

  $m_1=\frac{1}{2}$

  $(3,2)$

Here

  $x_1=3$

  $y_1=2$

Using point slope formula

  $y-3=-\frac{1}{2}(x-3)$

On solving

  $y-3=-\frac{1}{2}x+\frac{3}{2}$

Add 3 both sides

  $y-3+3=-\frac{1}{2}x+\frac{3}{2}+3$

On solving

  $y=-\frac{1}{2}x+\frac{9}{2}$

Hence, slope intercept form is $y=-\frac{1}{2}x+\frac{9}{2}$