WeBWorK using host: https://webwork.pcc.edu, format: simple, seed: 2032, course: orcca

$\newcommand{\subtractright}[1]{{\color{blue}{{{}-#1}}}}\newcommand{\divideunder}[2]{\frac{#1}{{\color{blue}{{#2}}}}}\newcommand{\amp}{&}$

Line $A$ and Line $B$ are perpendicular. Line $A$ ’s equation is $2x+3y=12\text{.}$ Line $B$ passes through the point $(4,-3)\text{.}$ Find an equation for Line $B\text{.}$

Solution

First, we will find Line $A$ ’s slope by rewriting its equation from standard form to slope-intercept form:

$\begin{aligned} 2x+3y\amp=12\\ 3y\amp=12\subtractright{2x}\\ 3y\amp=-2x+12\\ y\amp=\divideunder{-2x+12}{3}\\ y\amp=-\frac{2}{3}x+4 \end{aligned}$

So Line $A$ ’s slope is $-\frac{2}{3}\text{.}$ Since Line $B$ is perpendicular to Line $A\text{,}$ its slope is $-\frac{1}{-\frac{2}{3}}=\frac{3}{2}\text{.}$ It’s also given that Line $B$ passes through $(4,-3)\text{,}$ so we can write Line $B$ ’s point-slope form equation:

$\begin{aligned} y\amp=m(x-x_0)+y_0\\ y\amp=\frac{3}{2}(x-4)-3 \end{aligned}$