How do you write the equation of a line in slope intercept, point slope and standard form given (-3,3) and (2,3)?

May 2, 2017

See the solution explanation below:

Explanation:

Because the $y$ value for both points is $3$ by definition this is a horizontal line.

Horizontal lines have the equation $y = a$ where $a$ is the same value for $y$ for all values of $x$ and the slope is $m = 0$. For this problem the equation is:

$y = 3$

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 gives:

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

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

Substituting gives:

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

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 gives:

$\left(y - \textcolor{red}{3}\right) = \textcolor{b l u e}{0} \left(x - \textcolor{red}{c}\right)$

Where $c$ is any value.