# How do you find the equation of the line that passes through (5, -3), in Standard Form, and is parallel to the line that passes through (4, -5) and (-1, -6)?

Aug 30, 2015

$- x + 5 y = - 20$

#### Explanation:

You know that you must find the equation of a line that passes through point $\left(5 , - 3\right)$, and that is parralel to another line that passes through two points, $\left(4 , - 5\right)$ and $\left(- 1 , - 6\right)$.

The important thing to realize here is that two parralel lines have the same slope, but different $y$-intercepts.

This means that you can use the two points given for the second line to determine its slope, which will be equal to the slope of the target line.

For a general form line that passes through two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$, the slope is equal to

$\textcolor{b l u e}{\text{slope} = m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}}$

The slope of the two lines will be

$m = \frac{- 6 - \left(- 5\right)}{- 1 - 4} = \frac{\left(- 1\right)}{\left(- 5\right)} = \frac{1}{5}$

You now have the slope of the target line, which means that you can use the point $\left(5 , - 3\right)$ to write its equation in point slope form

$\textcolor{b l u e}{y - {y}_{1} = m \cdot \left(x - {x}_{1}\right)}$

$y - \left(- 3\right) = \frac{1}{5} \cdot \left(x - 5\right)$

$y + 3 = \frac{1}{5} \left(x - 5\right)$

The standard form of a line is

$\textcolor{b l u e}{A x + B y = C}$

$y = \frac{1}{5} x - 1 - 3$
$- \frac{1}{5} x + y = - 4 \text{ }$ $\iff$ $\text{ } - x + 5 y = - 20$