## sammietaygreen attachment! one year ago one year ago

1. whpalmer4

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.

2. whpalmer4

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.

3. sammietaygreen

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

4. whpalmer4

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

5. whpalmer4

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

6. sammietaygreen

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

7. whpalmer4

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

8. sammietaygreen

thanks! :)

9. whpalmer4

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!

10. sammietaygreen

could you help on one more on this thread?