Chapter 1, Problem 10CT

In Problems $1-3$, use $P_1=\left(-1,3\right)$ and $P_2=\left(5,-1\right)$.

For the line $2x+3y=6$, find a line parallel to the given line containing the point $(1,-1)$. Also find a line perpendicular to the given line containing the point $(0,3)$.

To determine

The line parallel to the line $2x+3y=6$ and containing the point $\left(1,-1\right)$, and also the perpendicular line containing the point $\left(0,3\right)$.

Answer to Problem 10CT

Solution:

The equation of the line parallel to line $2x+3y=6$ and containing the point $\left(1,-1\right)$ is $2x+3y=-1$, and the equation of the perpendicular line containing the point $\left(0,3\right)$ is $3x-2y=-6$.

Explanation of Solution

Given Information:

The equation of the line $2x+3y=6$, and the line parallel to it contains the point $\left(1,-1\right)$, and the perpendicular line contains the point $\left(0,3\right)$.

Explanation:

First, write the equation of line $2x+3y=6$ in slope-intercept form $y=mx+b$ to find its slope

$2x+3y=6$.

Subtracting $-2x$ from both sides,

$\Rightarrow 2x+3y-2x=-2x+6$,

$\Rightarrow 3y=-2x+6$.

Dividing by 3 on both sides,

$\Rightarrow \frac{3y}{3}=\frac{-2}{3}x+\frac{6}{3}$,

$\Rightarrow y=\frac{-2}{3}x+2$.

Thus, the slope intercept form of the line $2x+3y=6$ is $y=\frac{-2}{3}x+2$.

By comparing equation $y=\frac{-2}{3}x+2$ with $y=mx+b$, the slope of the line is $-\frac{2}{3}$.

The slope of the parallel lines is equal, so the slope of the line parallel to $2x+3y=6$ is $-\frac{2}{3}$.

Since the line contains point $\left(1,-1\right)$ and has slope $-\frac{2}{3}$,

Use point-slope form to get the equation of line

$y-y_1=m\left(x-x_1\right)$.

By substituting values of $m,x_1$ and $y_1$, it gives

$y-\left(-1\right)=-\frac{2}{3}\left(x-1\right)$,

$\Rightarrow 3\left(y+1\right)=-2\left(x-1\right)$,

$\Rightarrow 3y+3=-2x+2$,

$\Rightarrow 2x+3y=-1$.

Thus, the equation of the line parallel to line $2x+3y=6$, and containing point $\left(1,-1\right)$, is $2x+3y=-1$.

Now, to find the equation of the line perpendicular to line $2x+3y=6$ and containing the point $\left(0,3\right)$.

The product of the slopes of perpendicular lines is $-1$.

Hence, the slope of the line perpendicular to $2x+3y=6$ is $\frac{3}{2}$, since $-\frac{2}{3}\left(\frac{3}{2}\right)=-1$.

Now, find the equation of the line containing point $\left(0,3\right)$ and having slope $\frac{3}{2}$.

By using point slope form:

$y-3=\frac{3}{2}\left(x-0\right)$,

$\Rightarrow 2\left(y-3\right)=3x$,

$\Rightarrow 2y-6=3x$,

$\Rightarrow 3x-2y=-6$,

Thus, the equation of the line perpendicular to line $2x+3y=6$, and containing point $\left(0,3\right)$, is $3x-2y=-6$.