# How do you write an equation in standard form for a line passing through (3, 4) and is parallel to the line y = -x + 6?

May 2, 2015

Answer: $x + y = 7$

Explanation:

The standard form equation for a line on the $x y$-plane is $A x + B y = C$, where $A , B , C$ are constants. However, this problem may be easier if we begin in slope-intercept form (i.e. $y = m x + b$), since the equation of the parallel line is given in said form.

For two lines to be parallel in the $x y$-plane, they must have the same slope (represented by $m$ in slope-intercept form) and different Y-intercepts. The line parallel to our desired line is given the equation $y = - x + 6$, which could be rewritten as $y = \left(- 1\right) x + 6$ due to the identity property of multiplication. Our parallel line thus has a slope of $m = - 1$, meaning that our desired line possesses the same slope.

At this point, we may construct our desired line in either slope-intercept or (by skipping a step) standard form. We will begin with slope-intercept and then convert to standard form. Our desired line, given our discovery above, has an equation in slope-intercept form of $y = - x + b$, where $b$ is an unknown Y-intercept. We are given the coordinates for a point on the line $\left(3 , 4\right)$, and may substitute these coordinates in for $x$ and $y$ to determine $b$:
$y = - x + b \to 4 = - 3 + b \to 7 = b$

Having determined the Y-intercept, we now know that our line has a slope-intercept equation of $y = - x + 7$. By adding $x$ to either side-

$y = - x + 7 \to x + y = 7$

-we obtain the standard form equation of our line.