# How do you determine whether the lines for each pair of equations 3x+2y=-5 y=-2/3x+6 are parallel, perpendicular, or neither?

Mar 4, 2018

The lines are not parallell, nor are they perpendicular.

#### Explanation:

First, we get the two linear equations into $y = m x + b$ form:

${L}_{1} : y = - \frac{2}{3} x + 6 \to m = - \frac{2}{3}$

${L}_{2} : 3 x + 2 y = - 5$

${L}_{2} : 2 y = - 3 x - 5$

${L}_{2} : y = - \frac{3}{2} x - 5 \to m = - \frac{3}{2}$

If the lines were parallell, they would have the same $m$-value, which they don't, so they cannot be parallell.

If the two lines are perpendicular, their $m$-values would be negative reciprocals of each other. In the case of ${L}_{1}$, the negative reciprocal would be:
$- \frac{1}{- \frac{2}{3}} = - \left(- \frac{3}{2}\right) = \frac{3}{2}$

This is almost the negative reciprocal, but we're off by a minus sign, so the lines are not perpendicular.

Mar 4, 2018

Neither parallel nor perpendicular

#### Explanation:

Rearranging the $1$ st equation as $y = m x + c$,we get,

$y = - \frac{3}{2} x - \left(\frac{5}{2}\right)$ hence, slope =$- \frac{3}{2}$

the other equation is, $y = - \frac{2}{3} x + 6$ ,slope is $- \frac{2}{3}$

Now,slope of both the equations are not equal,so they are not parallel lines.

Again,product of their slope is $- \frac{3}{2} \cdot \left(- \frac{2}{3}\right) = 1$

But,for two lines to be perpendicular, product of their slope has to be $- 1$

So,they are not perpendicular as well.