# How do you write an equation of a line that goes through (6,-2); m= -4/3 in standard form?

May 12, 2015

Standard form is $y = m x + c$ where $m$ is the slope and $c$ is the $y$ coordinate of the intersection of the line with the $y$ axis.

We are given $m$, so it remains to determine $c$.

Subtracting $m x$ from both sides of the formula yields $c = y - m x$.

Given that the line passes through (6, -2) all we need do is substitute $x = 6$ and $y = - 2$ to get

$c = y - m x$

$= - 2 - \left(- \frac{4}{3}\right) \cdot 6 = - 2 + 4 \cdot \frac{6}{3} = - 2 + 8 = 6$

So the standard form is $y = - \left(\frac{4}{3}\right) x + 6$