# What is an equation in slope-intercept form of the line that is perpendicular to the graph of y=2x+3 and passes through (3, -4)?

Jun 13, 2018

$y = - \frac{1}{2} x - \frac{5}{2}$

#### Explanation:

The slope of the line perpendicular to the graph of $y = 2 x + 3$ is $- \frac{1}{2}$. The perpendicular slope is the negative inverse of the original slope. The product of perpendicular slopes is $- 1$, where:

${m}_{1} {m}_{2} = - 1$,

where:

${m}_{1}$ is the original slope $\left(2\right)$ and ${m}_{2}$ is the perpendicular slope.

$2 {m}_{2} = - 1$

Divide both sides by $2$.

${m}_{2} = - \frac{1}{2}$

So we now have the slope and we have been given a point $\left(\textcolor{red}{3} , \textcolor{b l u e}{- 4}\right)$.

Find the point-slope form of the perpendicular line.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Plug in the known values.

$y - \left(\textcolor{b l u e}{- 4}\right) = - \frac{1}{2} \left(x - \textcolor{red}{3}\right)$

$y + 4 = - \frac{1}{2} \left(x - 3\right)$ $\leftarrow$ point-slope form.

To convert the point-slope form to slope-intercept form, solve the point-slope form for $y$.

Slope-intercept form is: $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept.

$y + 4 = - \frac{1}{2} \left(x - 3\right)$

$y + 4 = - \frac{1}{2} x + \frac{3}{2}$

Subtract $4$ from both sides.j

$y = - \frac{1}{2} x + \frac{3}{2} - 4$

Multiply $4$ by $\frac{2}{2}$ to get an equivalent fraction with $2$ as the denominator.

$y = - \frac{1}{2} x + \frac{3}{2} - 4 \times \frac{2}{2}$

Simplify.

$y = - \frac{1}{2} x + \frac{3}{2} - \frac{8}{2}$

Simplify.

$y = - \frac{1}{2} x - \frac{5}{2}$ $\leftarrow$ perpendicular slope-intercept form

graph{(y-2x-3)(y+1/2x+5/2)=0 [-11.25, 11.25, -5.625, 5.625]}