# How do you write the equation of a line in slope intercept, point slope and standard form given (3,9) and (5,15)?

Feb 19, 2017

The point slope equations are: $\left(y - \textcolor{red}{9}\right) = \textcolor{b l u e}{3} \left(x - \textcolor{red}{3}\right)$ or $\left(y - \textcolor{red}{15}\right) = \textcolor{b l u e}{3} \left(x - \textcolor{red}{5}\right)$

The slope intercept form is: $y = \textcolor{red}{3} x + \textcolor{b l u e}{0}$

The standard for is: $\textcolor{red}{3} x + \textcolor{b l u e}{- 1} y = \textcolor{g r e e n}{0}$

#### 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:

$m = \frac{\textcolor{red}{15} - \textcolor{b l u e}{9}}{\textcolor{red}{5} - \textcolor{b l u e}{3}} = \frac{6}{2} = 3$

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:

$\left(y - \textcolor{red}{9}\right) = \textcolor{b l u e}{3} \left(x - \textcolor{red}{3}\right)$

We can also substitute the slope we calculated and the second point from the problem giving:

$\left(y - \textcolor{red}{15}\right) = \textcolor{b l u e}{3} \left(x - \textcolor{red}{5}\right)$

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. We can take the last equation and solve for $y$:

$y - \textcolor{red}{15} = \left(\textcolor{b l u e}{3} \times x\right) - \left(\textcolor{b l u e}{3} \times \textcolor{red}{5}\right)$

$y - \textcolor{red}{15} = 3 x - 15$

$y - \textcolor{red}{15} + 15 = 3 x - 15 + 15$

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

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

Transforming the last equation gives:

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

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

$- 3 x + y = 0$

$\textcolor{red}{- 1} \left(- 3 x + y\right) = \textcolor{red}{- 1} \times 0$

$\textcolor{red}{3} x + \textcolor{b l u e}{- 1} y = \textcolor{g r e e n}{0}$