# How do you the equation of the line that passes through point (-2, 3) and is parallel to the line formed by the equation y = 4x + 7?

Feb 15, 2017

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

Or

$y = 4 x + 11$

#### Explanation:

First let's find the slope of the equation we are looking for. We know it is parallel to the given line and so will have the same slope. The given line is 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.

So, for $y = \textcolor{red}{4} x + \textcolor{b l u e}{7}$ we know the slope is $\textcolor{red}{m = 4}$

We can now use the point-slope formula to find the equation of the line we are looking for. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$

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

Substituting the point from the problem and the slope we established gives:

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

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

Or, we can solve for $y$ to put this equation in slope-intercept form:

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

$y - \textcolor{red}{3} = 4 x + 8$

$y - \textcolor{red}{3} + 3 = 4 x + 8 + 3$

$y - 0 = 4 x + 11$

$y = 4 x + 11$