Chapter 4.3, Problem 32PPS

To determine

To Write: An equation of line in slope intercept form that passes through

$y-$ intercept of line $3x+2y=8$ that is also perpendicular to required line.

Answer to Problem 32PPS

An equation of line in slope- intercept form is $y=\frac{2}{3}x+4$

Explanation of Solution

Given information:

Line passes through $y-$ intercept and also perpendicular to line $3x+2y=8$

Concept and Formula Used:

Slope intercept form $y=mx+c$ where $m$ is slope and $c$ is intercept.

Point slope form of a line: $y-y_1=m\left(x-x_1\right)$

Perpendicular lines: If lines are perpendicular then the product of their slope is $-1$

$y-$ intercept of line is point where $x$ coordinate is zero.

Calculation:

The required line is perpendicular to

  $\begin{array}{l}3x+2y=8\\ 2y=-3x+8\\ y=-\frac{3}{2}x+4\end{array}$ .

First find the slope of given line by compairing with $y=mx+c$ .

The slope of line $3x+2y=8$ is $-\frac{3}{2}$ .

Let the slope of required line be ‘m’. As the required line is perpendicular to $3x+2y=8$

Therefore, product of slopes is $-1$

  $\begin{array}{l}m\times -\frac{3}{2}=-1\\ m=\frac{2}{3}\end{array}$

The required line passes through $y-$ intercept of line $3x+2y=8$

Putting $x=0$ in

  $\begin{array}{l}3x+2y=8\\ 3\left(0\right)+2y=8\\ 2y=8\\ y=4\end{array}$

Therefore point from which line passes is $\left(0,4\right)$ and having slope $\frac{2}{3}$

Using point slope form of equation of line

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \text{where }\left(x_1,y_1\right)=\left(0,4\right)\&m=\frac{2}{3}\\ y-\left(4\right)=\frac{2}{3}\left(x-\left(0\right)\right)\\ y-4=\frac{2}{3}\left(x\right)\\ \text{Ass 4 to both sides}\\ y-4+4=\frac{2}{3}\left(x\right)+4\\ y=\frac{2}{3}x+4\end{array}$

Conclusion:

An equation of line in slope- intercept form is $y=\frac{2}{3}x+4$