## ValentinaT Group Title What is the length of the longest side of a triangle that has vertices at (-5, 2) (1,-6) and (1,2)? one year ago one year ago

1. v4xN0s Group Title

Use the distance formula

2. ValentinaT Group Title

Distance formula?

3. v4xN0s Group Title

LOL

4. v4xN0s Group Title

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

5. MathLegend Group Title

I would also suggest labeling your coordinates

6. v4xN0s Group Title

MathLegend ur not a math legend at all

7. MathLegend Group Title

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

8. MathLegend Group Title

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

9. ValentinaT Group Title

Thank you.

10. MathLegend Group Title

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

11. hba Group Title

|dw:1355599903259:dw|

12. MathLegend Group Title

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 Group Title

Do you understand so far @ValentinaT ?

14. MathLegend Group Title

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

15. MathLegend Group Title

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

16. ValentinaT Group Title

Yeah, I'm getting it, thank you.

17. ValentinaT Group Title

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

18. MathLegend Group Title

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

19. MathLegend Group Title

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

20. ValentinaT Group Title

Okay.

21. MathLegend Group Title

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

22. MathLegend Group Title

36+64 = 100

23. MathLegend Group Title

$\sqrt{100}$

24. ValentinaT Group Title

10

25. MathLegend Group Title

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

26. MathLegend Group Title

Good.

27. MathLegend Group Title

So side AB = 10

28. MathLegend Group Title

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

29. MathLegend Group Title

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

30. MathLegend Group Title

@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. ValentinaT Group Title

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

32. MathLegend Group Title

Remember we are not trying to find a slope.

33. MathLegend Group Title

We are looking for the distance between the vertices.

34. MathLegend Group Title

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

35. MathLegend Group Title

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

36. MathLegend Group Title

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

37. ValentinaT Group Title

Okay, 0 + 64?

38. MathLegend Group Title

Good

39. MathLegend Group Title

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

40. ValentinaT Group Title

So, 8 as the square root.

41. MathLegend Group Title

Yes, so side BC = 8

42. MathLegend Group Title

Now, try side AC

43. MathLegend Group Title

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

44. ValentinaT Group Title

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

45. MathLegend Group Title

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

46. ValentinaT Group Title

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

47. MathLegend Group Title

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

48. MathLegend Group Title

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

49. ValentinaT Group Title

Okay, thank you!