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

2 Answers
Mar 4, 2018

Answer:

The lines are not parallell, nor are they perpendicular.

Explanation:

First, we get the two linear equations into #y=mx+b# form:

#L_1: y=-2/3x+6 -> m=-2/3#

#L_2: 3x+2y=-5#

#L_2: 2y=-3x-5#

#L_2: y=-3/2x-5 -> m=-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:
#-1/(-2/3)=-(-3/2)=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

Answer:

Neither parallel nor perpendicular

Explanation:

Rearranging the #1# st equation as #y=mx+c#,we get,

#y=-3/2x -(5/2)# hence, slope =#-3/2#

the other equation is, #y=-2/3x+6# ,slope is #-2/3#

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

Again,product of their slope is #-3/2 * (-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.