Chapter 3.4, Problem 39PPS

To determine

To write in point-slope form for the line parallel to $y=-\frac{2}{3}x+1$ and contain

  $(6,2)$

Answer to Problem 39PPS

Slope intercept form

  $y=-\frac{2}{3}x+6$

Explanation of Solution

Given information:

Line parallel to $y=-\frac{2}{3}x+1$ and contain $(6,2)$

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

  $y=-\frac{2}{3}x+1$

Then slope of given line

  $m_2=-\frac{2}{3}$

When lines are parallel

Then

  $m_1=m_2$

Then slope of required line is

  $m_1=-\frac{2}{3}$

Then slope and point containing by required line

  $m_1=-\frac{2}{3}$

  $(6,2)$

Here

  $x_1=6$

  $y_1=2$

Using point slope formula

  $y-2=-\frac{2}{3}(x-6)$

On solving

  $y-2=-\frac{2}{3}x+4$

Add 2 both sides

  $y-2+2=-\frac{2}{3}x+4+2$

On solving

  $y=-\frac{2}{3}x+6$

Hence, slope intercept form is $y=-\frac{2}{3}x+6$