sammietaygreen Group Title attachment! one year ago one year ago

1. whpalmer4 Group Title

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 Group Title

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 Group Title

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

4. whpalmer4 Group Title

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

5. whpalmer4 Group Title

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

6. sammietaygreen Group Title

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 Group Title

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

8. sammietaygreen Group Title

thanks! :)

9. whpalmer4 Group Title

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 Group Title

could you help on one more on this thread?