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

Inbrunstlr 2022-10-08 Answered
Determine whether the lines for each pair of equations 3x+2y=-5 y=-2/3x+6 are parallel, perpendicular, or neither
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## Answers (2)

Mckenna Friedman
Answered 2022-10-09 Author has 10 answers
First, we get the two linear equations into y=mx+b form:
${L}_{1}:y=-\frac{2}{3}x+6\to m=-\frac{2}{3}$
${L}_{2}:3x+2y=-5$
${L}_{2}:2y=-3x-5$
${L}_{2}:y=-\frac{3}{2}x-5\to m=-\frac{3}{2}$
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.
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Denisse Fitzpatrick
Answered 2022-10-10 Author has 3 answers
Rearranging the 1 st equation as y=mx+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.
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