# How do you write an equation in standard form when given slope and point on line (1, 5), m = -3?

##### 1 Answer
Jul 28, 2017

See a solution process below:

#### Explanation:

First, we can use the point-slope formula to write 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 $\left(\textcolor{red}{{x}_{1} , {y}_{1}}\right)$ is a point the line passes through.

Substituting the slope and values from the point in the problem gives:

$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{- 3} \left(x - \textcolor{red}{1}\right)$

We can now solve for the Standard Form of the equation. 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

$y - \textcolor{red}{5} = \left(\textcolor{b l u e}{- 3} \times x\right) - \left(\textcolor{b l u e}{- 3} \times \textcolor{red}{1}\right)$

$y - \textcolor{red}{5} = - \textcolor{b l u e}{3} x - \left(- 3\right)$

$y - \textcolor{red}{5} = - \textcolor{b l u e}{3} x + 3$

$y - \textcolor{red}{5} + 5 = - \textcolor{b l u e}{3} x + 3 + 5$

$y - 0 = \textcolor{b l u e}{3} x + 8$

$3 x + y = 3 x + \textcolor{b l u e}{3} x + 8$

$3 x + y = 0 + 8$

$\textcolor{red}{3} x + \textcolor{b l u e}{1} y = \textcolor{g r e e n}{8}$