## anonymous one year ago Coordinates for D. (?,6) C is (10,0) Rectangle ABCD. A(2,14) B(-2,8) C(10,0) D(?,6)

1. anonymous

@jim_thompson5910

2. jim_thompson5910

earlier we found line BC to be $\Large y = -\frac{2}{3}x+\frac{20}{3}$

3. anonymous

Yes, would it be the same for AD?

4. jim_thompson5910

no

5. jim_thompson5910

but the same steps will be used find the slope of AD (since it's parallel to BC, the two slopes are equal). Then use it to find the equation of line AD

6. jim_thompson5910

you also need the equation for line CD

7. anonymous

For AD equation, would it start like: y-14=m(x-2)? m = slope

8. jim_thompson5910

yes and m = -2/3 because AD and BC have the same slope (they are parallel lines)

9. anonymous

So, then I can plug in 6 to get the x value for D?

10. jim_thompson5910

not quite

11. jim_thompson5910

you'll need equation CD

12. jim_thompson5910

oh wait, nvm

13. jim_thompson5910

yes you can plug in y = 6. I didn't see D(?,6)

14. anonymous

Yeah, sorry, with my packet, I have image of it showing the C point being on 0. So, helped me determine the distance for the AD y and what to subtract.

15. anonymous

-2/3 * -2 would be 12/3?

16. jim_thompson5910

-2/3 * -2 = 4/3

17. anonymous

6-14=-2/3(x-2) -8=-2/3x + 4/3 It ends up being: 6 = -2/3x + 28/3 right?

18. jim_thompson5910

good

19. jim_thompson5910

now solve for x

20. anonymous

x=5?

21. jim_thompson5910

no

22. jim_thompson5910

oh I see what went wrong

23. jim_thompson5910

When you went from -8=-2/3x + 4/3 to 6 = -2/3x + 28/3 you made an error the 6 in  6 = -2/3x + 28/3 should be 0 so you should have 0 = -2/3x + 28/3

24. jim_thompson5910

it's 0 because you added 8 to both sides the -8 and +8 combine to 0 on the left side

25. anonymous

OH. x=14

26. jim_thompson5910

yep D is (14,6)

27. anonymous

To be honest, I read my graph wrong aswell, so I was confused. I lack in graphing, but all is well. You really helped and taught me a few things, I hope to remember. (:

28. jim_thompson5910

I'm glad I was of help