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whpalmer4 · Answer

Put both equations in slope-intercept form: y = mx + b where m is slope and b is y-intercept.  If they are parallel, slopes will be =.  If they are perpendicular, the product of the two slopes will be -1.  Otherwise, they are none of the above.

whpalmer4 · Answer

Slope of first line is -2/3.  Second equation we need to solve for y:
[\2x-3y=-3\]$$2x+3=3y$$$$y=\frac{2}{3}x+1$$ so slope of second line is 2/3.

$$\frac{2}{3}*-\frac{2}{3} = -\frac{4}{9} $$ so they are not perpendicular.

sammietaygreen · Answer

Yes, yes. I know it can't be perpendicular.

whpalmer4 · Answer

Well, that also shows that they are not parallel because the slopes are not equal.

whpalmer4 · Answer

Need to check your work if you graphed them and got parallel lines :-)

sammietaygreen · Answer

Ohhh, I must have graphed wrong. That simplified it a bit, @jim_thompson5910 is helping me as well. I'm going to check what I did wrong

whpalmer4 · Answer

I always use x = 0 as one of my graph points :-)

sammietaygreen · Answer

thanks! :)

whpalmer4 · Answer

In this case, it doesn't help because they actually cross at x =0, but often that isn't the case, and the arithmetic is a bit easier!

sammietaygreen · Answer

could you help on one more on this thread?