# How do you find the slope of a line that is a) parallel to the line given and b)perpendicular to the line with the given equation 2x-5y = -7?

May 19, 2015

If a line is expressed by the equation $y = m x + c$ then $m$ is the slope and $c$ is the intercept.

Starting with the equation given, first add $5 y$ to both sides to get:

$2 x = 5 y - 7$

Add $7$ to both sides to get:

$2 x + 7 = 5 y$

Divide both sides by $5$ to get:

$\frac{2}{5} x + \frac{7}{5} = y$

In other words $y = \frac{2}{5} x + \frac{7}{5}$

Comparing with $y = m x + c$, we see that the slope is $\frac{2}{5}$ and the intercept is $\frac{7}{5}$.

Any line parallel to this line will have the same slope: $\frac{2}{5}$.

Any line perpendicular to a line of slope $\frac{2}{5}$ will have a slope of $- \frac{5}{2}$.

There are several ways to see this.

For example, consider the following steps:
(1) Reflect the original line in the line $y = x$. This will have the effect of swapping $x$ and $y$ in the original equation.
(2) Reflect the line in the $x$ axis, given by the equation $y = 0$. This will have the effect of inverting the sign of the $y$ coordinate.

The total effect of (1) followed by (2) is a rotation of $\frac{\pi}{2}$. That is, the resulting line is perpendicular to the line you started with.

Starting with your original line this will result in a line with equation $- 2 y - 5 x = - 7$, which can be reformulated as:
$y = - \left(\frac{5}{2}\right) x + \frac{7}{2}$. This has slope $- \frac{5}{2}$ as predicted.