# How do you write an equation in standard form given point (-2,-5) and (-1,3)?

Feb 4, 2017

$\textcolor{red}{8} x - \textcolor{b l u e}{1} y = \textcolor{g r e e n}{- 11}$

#### Explanation:

We can first write an equation in point-slope form. To do this we must first determine the slope. 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 in the problem gives:

$m = \frac{\textcolor{red}{3} - \textcolor{b l u e}{- 5}}{\textcolor{red}{- 1} - \textcolor{b l u e}{- 2}}$

$m = \frac{\textcolor{red}{3} + \textcolor{b l u e}{5}}{\textcolor{red}{- 1} + \textcolor{b l u e}{2}}$

$m = \frac{8}{1} = 8$

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 gives:

$\left(y - \textcolor{red}{- 5}\right) = \textcolor{b l u e}{8} \left(x - \textcolor{red}{- 2}\right)$

$\left(y + \textcolor{red}{5}\right) = \textcolor{b l u e}{8} \left(x + \textcolor{red}{2}\right)$

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

We can now transform the point-slope form to the standard form:

$y + \textcolor{red}{5} = \left(\textcolor{b l u e}{8} \times x\right) + \left(\textcolor{b l u e}{8} \times \textcolor{red}{2}\right)$

$y + 5 = 8 x + 16$

$\textcolor{red}{- 8 x} + y + 5 - \textcolor{b l u e}{5} = \textcolor{red}{- 8 x} + 8 x + 16 - \textcolor{b l u e}{5}$

$- 8 x + y + 0 = 0 + 11$

$- 8 x + y = 11$

$\textcolor{red}{- 1} \left(- 8 x + y\right) = \textcolor{red}{- 1} \times 11$

$8 x - y = - 11$