## anonymous 3 years ago What is the length of the longest side of a triangle that has vertices at (-5, 2) (1,-6) and (1,2)?

1. anonymous

Use the distance formula

2. anonymous

Distance formula?

3. anonymous

LOL

4. anonymous

ok its $d=\sqrt{(x _{1}-x _{2})^2+(y _{1}-y _{2})^2}$

5. MathLegend

I would also suggest labeling your coordinates

6. anonymous

MathLegend ur not a math legend at all

7. MathLegend

A= (-5, 2) B = (1,-6) C = (1,2)

8. MathLegend

So when using the distance formula... you can solve for side AB first.

9. anonymous

Thank you.

10. MathLegend

So using that formula, can you label the coordinates as (x1, y1) & (x2,y2)

11. hba

|dw:1355599903259:dw|

12. MathLegend

A= (-5, 2) B = (1,-6) So lets solve for side AB first... so, "A" comes first right? So let that be your x1 & y1 Then, "B" can be your x2 & y2

13. MathLegend

Do you understand so far @ValentinaT ?

14. MathLegend

All I did was label them so that we can plug it into that distance formula.

15. MathLegend

@ValentinaT let me know when you get back so we can work this out together. :)

16. anonymous

Yeah, I'm getting it, thank you.

17. anonymous

$\frac{ -6 - 2 }{ -5 - 1 } = \frac{ -8 }{ -6}$

18. MathLegend

So for AB (1+5)^2+(-6-2)^2

19. MathLegend

I'm sorry the formula is actually x2-x1 and y2-y1 the above poster just mixed up the two.

20. anonymous

Okay.

21. MathLegend

(1+5)^2+(-6-2)^2 (6)^2+(-8)^2 36+64

22. MathLegend

36+64 = 100

23. MathLegend

$\sqrt{100}$

24. anonymous

10

25. MathLegend

Now, we need the square root... because if you notice that entire formula had the square root symbol over it

26. MathLegend

Good.

27. MathLegend

So side AB = 10

28. MathLegend

So that is one side. So lets go for side BC

29. MathLegend

B = (1,-6) x1 y1 C = (1,2) x2 y2

30. MathLegend

@ValentinaT do you feel comfortable trying it out on your own? Tell me what you get and I'll check to see if your answer for side BC is correct.

31. anonymous

$\frac{ 2 - -6 }{ 1 - 1 } = \frac{ 8 }{ 0 }$

32. MathLegend

Remember we are not trying to find a slope.

33. MathLegend

We are looking for the distance between the vertices.

34. MathLegend

(1-1)^2 + (2+6)^2

35. MathLegend

To get that all I did was plug it into the formula. (x2-x1)^2+(y2-y1)^2

36. MathLegend

(1-1)^2 + (2+6)^2 Try solving that.

37. anonymous

Okay, 0 + 64?

38. MathLegend

Good

39. MathLegend

0 + 64 = 64 $\sqrt{64}$

40. anonymous

So, 8 as the square root.

41. MathLegend

Yes, so side BC = 8

42. MathLegend

Now, try side AC

43. MathLegend

(x2-x1)^2+(y2-y1)^2

44. anonymous

A = -5, 2 = x1, y1 C = 1, 2 = x2, y2 $\frac{ 1 - -5^2}{ 2 - 2^2 } = \frac{ 6 }{ 1 }$ so 36?

45. MathLegend

(x2-x1)^2+(y2-y1)^2 (1+5)^2 + (2-2)^2 Do you understand this step?

46. anonymous

36 + 0 Yeah I get it, I just couldn't figure out how to write it.

47. MathLegend

Good so if you took the square root of 36 $\sqrt{36}$

48. MathLegend

So now, we know the length of the longest side is AB

49. anonymous

Okay, thank you!