## anonymous one year ago ∆ABC has vertices A(0, 0), B(2, 4), and C(4, 2). ∆RST has vertices R(0, 3), S(-1, 5), and T(-2, 4). Prove that ∆ABC ∼ ∆RST. (Hint: Graph ∆ABC and ∆RST in the coordinate plane.)

1. dm91

2. steve816

Well, do you know how to plot points on the coordinate plane? All you do is draw the points and connect the dots.

3. dm91

|dw:1443590087569:dw|

4. anonymous

omg thank you

5. steve816

No problem :)

6. anonymous

so how do i do this problem?

7. anonymous

@steve816

8. dm91

ok so as steve put it do you have a graph?

9. anonymous

ive already graphed the points but i dont know how to prove it similar

10. steve816

If the triangles are the same size once you graphed, then they are similar.

11. anonymous

i thought you had to do the distance formula?

12. anonymous

they werent the same size

13. steve816

Yes, you can use the distance formula

14. steve816

For each of the sides

15. anonymous

okay how can i do it with the distance formula?

16. steve816

Plug in your (x, y) coordinates into this formula: http://2.bp.blogspot.com/-GMrbM3fg4NU/VHljne7DleI/AAAAAAAABec/3emecyWG4ew/s1600/Slide1.jpg

17. anonymous

ok give me a second

18. anonymous

wait which numbers do i put into it?

19. anonymous

i did AB and got 20 is that correct?

20. anonymous

?

21. anonymous

@steve816 is that correct?

22. anonymous

hello?

23. anonymous

i need help fast can someone help?

24. dm91

ok now what?

25. anonymous

is that correct?

26. anonymous

i got the squareroot of 20 for AB

27. dm91

yes

28. anonymous

okay thanks

29. dm91

yep

30. anonymous

for rs i got the sqaureroot of 5 is that correct?

31. dm91

yes

32. anonymous

and for CT i got the sqareroot of 40 is that correct?

33. dm91

yes

34. anonymous

ok what do i do now?

35. dm91

no that one would have to be up to Steve because he gave you the square root idea but i might be able to get you someone else that can help because Steve is helping other pp at the time.

36. anonymous

okay thanks!

37. dm91

@dan815

38. dm91

yep

39. dm91

@Koikkara

40. Koikkara

41. anonymous

i was doing the distance formula for the problem and dont know what to do next

42. Koikkara

43. anonymous

yes

44. anonymous

i did Ab and got the squareroot of 20 then i did RS and got the sqareroot of 5 then i did CT and got the squareroot of 40

45. Koikkara

this is cheating a bit becasue this program gives the lentgh of the sides but you need to use the SSS postulate and this is easier because 2 side length are the same $\frac{ BC}{ EF} = \frac{ AB }{DF}$