# How do you find a standard form equation for the line with (-5, 4) and (2, -5)?

##### 1 Answer
Jan 29, 2017

$\textcolor{red}{9} x + \textcolor{b l u e}{7} y = \textcolor{g r e e n}{- 17}$

#### Explanation:

First, we find the slope of the line.

The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$

Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.

Substituting the values from the points given in the problem produces:

$m = \frac{\textcolor{red}{- 5} - \textcolor{b l u e}{4}}{\textcolor{red}{2} - \textcolor{b l u e}{- 5}}$

$m = \frac{\textcolor{red}{- 5} - \textcolor{b l u e}{4}}{\textcolor{red}{2} + \textcolor{b l u e}{5}}$

$m = - \frac{9}{7}$

We can now use the point-slope formula to find an equation for the line.

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 slope we calculated and the first point from the problem gives:

$\left(y - \textcolor{red}{4}\right) = \textcolor{b l u e}{- \frac{9}{7}} \left(x - \textcolor{red}{- 5}\right)$

$\left(y - \textcolor{red}{4}\right) = \textcolor{b l u e}{- \frac{9}{7}} \left(x + \textcolor{red}{5}\right)$

Now, we must transform this equation to standard form.

The standard form of a linear equation is:

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

where, if at all possible, $\textcolor{red}{A}$, $\textcolor{b l u e}{B}$, and $\textcolor{g r e e n}{C}$are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

First, multiply both sides of the equation by $\textcolor{red}{7}$ to eliminate the fraction:

$\textcolor{red}{7} \left(y - 4\right) = \textcolor{red}{7} \times - \frac{9}{7} \left(x + 5\right)$

$7 y - 28 = \cancel{\textcolor{red}{7}} \times - \frac{9}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} \left(x + 5\right)$

7y - 28= -9((x + 5)

$7 y - 28 = - 9 x - 45$

$\textcolor{red}{9 x} + 7 y - 28 + \textcolor{b l u e}{28} = \textcolor{red}{9 x} + - 9 x - 45 + \textcolor{b l u e}{28}$

$9 x + 7 y - 0 = 0 - 45 + 28$

$9 x + 7 y = - 17$