Chapter 3.4, Problem 5CYP

To determine

To write in point-slope form for the line parallel to $y=-\frac{3}{4}x+3$ and contain $(-3,6)$

Answer to Problem 5CYP

Slope intercept form

  $y=-\frac{3}{4}x+\frac{15}{4}$

Explanation of Solution

Given information:

Line parallel to $y=-\frac{3}{4}x+3$ and contain $(-3,6)$

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

  $y=-\frac{3}{4}x+3$

Then slope of given line

  $m_2=-\frac{3}{4}$

When lines are parallel

Then

  $m_1=m_2$

Then slope of required line is

  $m_1=-\frac{3}{4}$

Point containing $(-3,6)$

  $x_1=-3$

  $y_1=6$

Using point slope formula

  $y-6=-\frac{3}{4}(x-(-3))$

On solving

  $y-6=-\frac{3}{4}(x+3)$

On solving

  $y-6=-\frac{3}{4}x-\frac{9}{4}$

Add 6 both sides

  $y-6+6=-\frac{3}{4}x-\frac{9}{4}+6$

On solving

  $y=-\frac{3}{4}x+\frac{15}{4}$

Hence, slope intercept form is $y=-\frac{3}{4}x+\frac{15}{4}$