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
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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:

$-\frac{1}{-\frac{2}{3}}=-(-\frac{3}{2})=\frac{3}{2}$

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

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

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

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

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

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:

$-\frac{1}{-\frac{2}{3}}=-(-\frac{3}{2})=\frac{3}{2}$

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

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 $-\frac{3}{2}\cdot (-\frac{2}{3})=1$

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

So,they are not perpendicular as well.

$y=-\frac{3}{2}x-\left(\frac{5}{2}\right)$ 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 $-\frac{3}{2}\cdot (-\frac{2}{3})=1$

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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