# How do you find a standard form equation for the line with (0, -10) and (-4, 0)?

Jul 14, 2017

See a solution process below:

#### Explanation:

First, we need to 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:

$m = \frac{\textcolor{red}{0} - \textcolor{b l u e}{- 10}}{\textcolor{red}{- 4} - \textcolor{b l u e}{0}} = \frac{\textcolor{red}{0} + \textcolor{b l u e}{10}}{\textcolor{red}{- 4} - \textcolor{b l u e}{0}} = \frac{10}{-} 4 = - \frac{5}{2}$

Because the first point in the problem has a value of $0$ for $x$ we know this is the $y$-intercept. The y-intercept is therefore: $\textcolor{b l u e}{- 10}$

We can now write and equation in slope-intercept form. The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

Substituting the slope we calculated and the $y$-intercept we determined gives:

$y = \textcolor{red}{- \frac{5}{2}} x + \textcolor{b l u e}{- 10}$

$y = \textcolor{red}{- \frac{5}{2}} x - \textcolor{b l u e}{10}$

We need to convert this equation to the Standard Form for Linear equations. 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, we can add $\frac{5}{2} x$ to each side of the equation to put the $x$ and $y$ variables on the left side of the equation as required by the Standard formula:

$\frac{5}{2} x + y = \frac{5}{2} x + \textcolor{red}{- \frac{5}{2}} x - \textcolor{b l u e}{10}$

$\frac{5}{2} x + y = 0 - \textcolor{b l u e}{10}$

$\frac{5}{2} x + y = - 10$

We can now multiply each side of the equation by $\textcolor{red}{2}$ to eliminate the fraction as required by the Standard formula:

$\textcolor{red}{2} \left(\frac{5}{2} x + y\right) = \textcolor{red}{2} \times - 10$

$\left(\textcolor{red}{2} \times \frac{5}{2} x\right) + \left(\textcolor{red}{2} \times y\right) = - 20$

$\left(\cancel{\textcolor{red}{2}} \times \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} x\right) + 2 y = - 20$

$\textcolor{red}{5} x + \textcolor{b l u e}{2} y = \textcolor{g r e e n}{- 20}$