Chapter 3.4, Problem 18PPS

To determine

To write in slope intercept form for the line that having slope $\frac{5}{11}$ and containing $(0,-3)$ .Then graph the line.

Answer to Problem 18PPS

Line

  $y=\frac{5}{11}x-3$

Then graph

Explanation of Solution

Given information:

The line having slope and point

  $m:\frac{5}{11}$

  $(0,-3)$

Formula used:

Point-slope form

  $y-y_1=m(x-x_1\text{)}$

Calculation:

The line having slope and point

  $m:\frac{5}{11}$

  $(0,-3)$

Here

  $x_1=0$

  $y_1=-3$

Substitute value of slope and point in the point-slope form

  $y-(-3)=\frac{5}{11}(x-0\text{) }$

On solving

  $y+3=\frac{5}{11}x$

Subtract 3 both sides

  $y+3-3=\frac{5}{11}x-3$

On solving

  $y=\frac{5}{11}x-3$

Substituting $x=0$

Then

  $y=-3$

Substituting $y=0$

Then

  $\text{0}=\frac{5}{11}x-3$

On solving

  $\frac{5}{11}x=3$

On solving

  $x=\frac{33}{5}$

Table

Then graph