# How do you find a standard form equation for the line with P (5, 5), and Q (-3, -3)?

Apr 24, 2017

See the entire solution process below:

#### Explanation:

First, we must determine 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 in the problem gives:

$m = \frac{\textcolor{red}{- 3} - \textcolor{b l u e}{5}}{\textcolor{red}{- 3} - \textcolor{b l u e}{5}} = - \frac{8}{-} 8 = 1$

Next, we can use the point-slope formula to get 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:

$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{1} \left(x - \textcolor{red}{5}\right)$

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 equation we wrote to standard form as follows:

$y - \textcolor{red}{5} = x - \textcolor{red}{5}$

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

$- x + y - 0 = 0 - 0$

$- x + y = 0$

$\textcolor{red}{- 1} \left(- x + y\right) = \textcolor{red}{- 1} \cdot 0$

$x - y = 0$

$\textcolor{red}{1} x + \textcolor{b l u e}{- 1} y = \textcolor{g r e e n}{0}$