# How do you write an equation in standard form for a line passing through (–4, 2) and is parallel to y = –x + 6?

May 1, 2015

All lines parallel to $y = a x + b$ have to have the same coefficient at $x$ in the form $y = p x + q$. That is, $p = a$.

In our case all lines parallel to $y = - x + 6$ have to have a form $y = - x + q$, where $q$ can be any real number.

The condition that our line should pass through $\left(- 4 , 2\right)$ results in the equation, from which we can determine an unknown variable $q$:
$2 = - \left(- 4\right) + q$
$q = - 2$

Therefore, the equation of a line we are looking for is
$y = - x - 2$

It passes through $\left(- 4 , 2\right)$ since, if we substitute $- 4$ for $x$, the value of $y$ will be $2$.
It also parallel to $y = - x + 6$ because the coefficients at $x$ are the same in both cases.

Graphs of these two equations are:
$y = - x + 6$
graph{-x+6 [-10, 10, -5, 5]}
$y = - x - 2$ - notice the line is passing point $\left(- 4 , 2\right)$ and is parallel to $y = - x + 6$ above
graph{-x-2 [-10, 10, -5, 5]}