Chapter 1, Problem 7CT

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

Write the slope-intercept form of the line with slope -2 containing the point $(3,-4)$. Graph the line.

To determine

The slope-intercept form of the line containing the point $\left(3,-4\right)$ and having slope $-2$, and graph the line.

Answer to Problem 7CT

Solution:

The slope-intercept form of the line is $y=-2x+2$.

The graph of the line $y=-2x+2$ is:

Explanation of Solution

Given Information:

The point is $\left(3,-4\right)$, and the slope of line is $-2$.

Explanation:

Point-slope form of an equation of a line is $y-y_1=m\left(x-x_1\right)$.

Here, $m=-2$, and $\left(x_1,y_1\right)=\left(3,-4\right)$.

By substituting values of $m,x_1$ and $y_1$ in $y-y_1=m\left(x-x_1\right)$, it gives

$y-\left(-4\right)=-2\left(x-3\right)$.

By simplifying,

$\Rightarrow y+4=-2x+6$.

$\Rightarrow y=-2x+6-4$.

$\Rightarrow y=-2x+6-4$.

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

So, the slope-intercept form of the line containing point $\left(3,-4\right)$, and having slope $-2$, is $y=-2x+2$.

The graph of the equation of line $y=-2x+2$ is: