# What is the point-slope form of the three lines that pass through (1,-2), (5,-6), and (0,0)?

##### 1 Answer
Aug 30, 2017

See a solution process below:

#### Explanation:

First, let's name the three points.

$A$ is $\left(1 , - 2\right)$; $B$ is $\left(5 , - 6\right)$; $C$ is $\left(0 , 0\right)$

First, let's find the slope of each 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.

Slope A-B:

${m}_{A - B} = \frac{\textcolor{red}{- 6} - \textcolor{b l u e}{- 2}}{\textcolor{red}{5} - \textcolor{b l u e}{1}} = \frac{\textcolor{red}{- 6} + \textcolor{b l u e}{2}}{\textcolor{red}{5} - \textcolor{b l u e}{1}} = - \frac{4}{4} = - 1$

Slope A-C:

${m}_{A - C} = \frac{\textcolor{red}{0} - \textcolor{b l u e}{- 2}}{\textcolor{red}{0} - \textcolor{b l u e}{1}} = \frac{\textcolor{red}{0} + \textcolor{b l u e}{2}}{\textcolor{red}{0} - \textcolor{b l u e}{1}} = \frac{2}{-} 1 = - 2$

Slope B-C:

${m}_{A - B} = \frac{\textcolor{red}{0} - \textcolor{b l u e}{- 6}}{\textcolor{red}{0} - \textcolor{b l u e}{5}} = \frac{\textcolor{red}{0} + \textcolor{b l u e}{6}}{\textcolor{red}{0} - \textcolor{b l u e}{5}} = \frac{6}{-} 5 = - \frac{6}{5}$

The point-slope form of a linear equation is: $\left(y - \textcolor{b l u e}{{y}_{1}}\right) = \textcolor{red}{m} \left(x - \textcolor{b l u e}{{x}_{1}}\right)$

Where $\left(\textcolor{b l u e}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ is a point on the line and $\textcolor{red}{m}$ is the slope.

We can substitute each of the slopes we calculated and one point from each line to write an equation in point-slope form:

Line A-B:

$\left(y - \textcolor{b l u e}{- 2}\right) = \textcolor{red}{- 1} \left(x - \textcolor{b l u e}{1}\right)$

$\left(y + \textcolor{b l u e}{2}\right) = \textcolor{red}{- 1} \left(x - \textcolor{b l u e}{1}\right)$

Or

$\left(y + \textcolor{b l u e}{2}\right) = \textcolor{red}{-} \left(x - \textcolor{b l u e}{1}\right)$

Line A-C:

$\left(y - \textcolor{b l u e}{- 2}\right) = \textcolor{red}{- 2} \left(x - \textcolor{b l u e}{1}\right)$

$\left(y + \textcolor{b l u e}{2}\right) = \textcolor{red}{- 2} \left(x - \textcolor{b l u e}{1}\right)$

Line B-C:

$\left(y - \textcolor{b l u e}{- 6}\right) = \textcolor{red}{- \frac{6}{5}} \left(x - \textcolor{b l u e}{5}\right)$

$\left(y + \textcolor{b l u e}{6}\right) = \textcolor{red}{- \frac{6}{5}} \left(x - \textcolor{b l u e}{5}\right)$