# How do you write the equation of the line passing through (-4,-5) with slope -3/2?

Jun 16, 2018

See a solution process below:

#### Explanation:

The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.

Substituting the values from the point and slope in the problem gives:

$\left(y - \textcolor{b l u e}{- 5}\right) = \textcolor{red}{- \frac{3}{2}} \left(x - \textcolor{b l u e}{- 4}\right)$

$\left(y + \textcolor{b l u e}{5}\right) = \textcolor{red}{- \frac{3}{2}} \left(x + \textcolor{b l u e}{4}\right)$

We can also solve for $y$ to put the equation in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

$y + \textcolor{b l u e}{5} = \left(\textcolor{red}{- \frac{3}{2}} \times x\right) + \left(\textcolor{red}{- \frac{3}{2}} \times \textcolor{b l u e}{4}\right)$

$y + \textcolor{b l u e}{5} = \textcolor{red}{- \frac{3}{2}} x + \left(- \frac{12}{2}\right)$

$y + \textcolor{b l u e}{5} = \textcolor{red}{- \frac{3}{2}} x + \left(- 6\right)$

$y + \textcolor{b l u e}{5} = \textcolor{red}{- \frac{3}{2}} x - 6$

$y + \textcolor{b l u e}{5} - 5 = \textcolor{red}{- \frac{3}{2}} x - 6 - 5$

$y + 0 = \textcolor{red}{- \frac{3}{2}} x - 11$

$y = \textcolor{red}{- \frac{3}{2}} x - \textcolor{b l u e}{11}$