Chapter 4.4, Problem 32PPS

To determine

To find: The equation of the straight line in slope-intercept form which passes through the point y-intercept of line $3x+2y=8$ and perpendicular to graph of same line.

Answer to Problem 32PPS

Equation of line is $y=\frac{2}{3}x+4$ .

Explanation of Solution

Given information: The given line is $3x+2y=8$ . And, the perpendicular line to this line passes through x-intercept of this line.

Calculation: Slope $\left(m\right)$ of the line perpendicular to line $3x+2y=8$ will be $\left(m=\frac{2}{3}\right)$ .

Now, y-intercept of line $3x+2y=8$ can be obtained by plugging $x=0$ in $3x+2y=8$ .

  $\begin{array}{c}3\left(0\right)+2y=\mathrm{8......}\left(\text{For }x=0\right)\\ 2y=8\\ y=4\end{array}$

Now, y-intercept of line $3x+2y=8$ is $\left(0,4\right)$ . Also the line perpendicular to this line passes through this point $\left(0,4\right)$ having slope $\left(m=\frac{2}{3}\right)$ . Therefore, the equation of line perpendicular to line $3x+2y=8$ can be written as,

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\mathrm{......}\left(\text{Point-slope form}\right)\\ y-\left(4\right)=\frac{2}{3}\left(x-0\right)\mathrm{......}\left(\begin{array}{l}\left(x_1,y_1\right)\to \left(0,4\right)\\ \text{and }m=\frac{2}{3}\end{array}\right)\\ y-4=\frac{2}{3}x\\ \text{or, }y=\frac{2}{3}x+4\end{array}$