## SephI 2 years ago Help with graphing, please :P

1. SephI

Choose the slope-intercept equation of the line that passes through the point (-5, -1) and is perpendicular to y = 5/2x + 2

2. SephI

They have the same slope since they are perpendicular. so I know (-5, -1) with a slope of 5/2 has something to do with the equation, but what happens next?

3. akaisha

they dont have the same slope, that is when they are parallel. For perpendicular the slope is opposite

4. akaisha

its a negative reciprocal, so can you figure it out what the slope is from there?

5. hba

6. SephI

Oh, ok. So I would have a slope of 2/5, and then what?

7. akaisha

close.. m=-2/5 then you use your point-interecpt form: y-y=m(x-x)

8. SephI

y - (-1) = -2/5[(x - (-5)] y + 1 = -2/5x + 5 y = -2/5x - 4? ^ That's not one of my options. So I did it wrong...

9. akaisha

thats close but you need to distribute the (-2/5) to the 5 as well

10. SephI

Why is that?

11. akaisha

that is because the x+5 is in parentheses. so the -2/5 applies to both pieces

12. akaisha

or (x-(-5)) however you want to think about it

13. SephI

y - (-1) = -2/5[(x - (-5)] y - (-1) = -2/5x + ??? How would I distribute the 5? Wouldn't it be -2/5*5 Which is -10/5 = -2?

14. akaisha

yes! perfect. so you would have (y-(-1))=-2/5x-2 then you just need to y by itself

15. SephI

y + 1 = -2/5x - 2 y = -2/5x - 3 - 1 y = -2/5x - 3?

16. akaisha