SOLUTION: Give triangele RST with vertices R(-2,1) S(-1,3) and T(2,1) RS equals ? I think the square root of 5 but Im not sure if I approched it correectly.

Algebra ->  Triangles -> SOLUTION: Give triangele RST with vertices R(-2,1) S(-1,3) and T(2,1) RS equals ? I think the square root of 5 but Im not sure if I approched it correectly.      Log On


   

Question 256137: Give triangele RST with vertices R(-2,1) S(-1,3) and T(2,1) RS equals ?
I think the square root of 5 but Im not sure if I approched it correectly.
Found 3 solutions by richwmiller, jim_thompson5910, checkley77:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Why do you have a question mark? There is no question.
I like all square roots. And round ones too!

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
We basically want to find the distance from R(-2,1) to S(-1,3)


Note: is the first point . So this means that x%5B1%5D=-2 and y%5B1%5D=1.
Also, is the second point . So this means that x%5B2%5D=-1 and y%5B2%5D=3.


d=sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29 Start with the distance formula.


d=sqrt%28%28-2--1%29%5E2%2B%281-3%29%5E2%29 Plug in x%5B1%5D=-2, x%5B2%5D=-1, y%5B1%5D=1, and y%5B2%5D=3.


d=sqrt%28%28-1%29%5E2%2B%281-3%29%5E2%29 Subtract -1 from -2 to get -1.


d=sqrt%28%28-1%29%5E2%2B%28-2%29%5E2%29 Subtract 3 from 1 to get -2.


d=sqrt%281%2B%28-2%29%5E2%29 Square -1 to get 1.


d=sqrt%281%2B4%29 Square -2 to get 4.


d=sqrt%285%29 Add 1 to 4 to get 5.


So our answer is d=sqrt%285%29 which means that the length of RS is sqrt%285%29 units.

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
R(-2,1) S(-1,3) and T(2,1)
d(rs)=sqrt[(x2-x1)^2+(y2-y1)^2]
d(rs)=sqrt[(-1+2)^2+(3-1)^2]
d(rs)=sqrt[1^2+2^2]
d(rs)=sqrt[1+4]
d(rs)=sqrt5
d(rs)=2.236 ans.