SOLUTION: I'm trying to solve this parallelogram I'm given 4 points and need to solve by using the distance formula. The 4 points are G(-1,3), H(2,1), F(3,0), and E(0,-2). I'm trying

Question 1108834: I'm trying to solve this parallelogram I'm given 4 points and need to solve by using the distance formula.
The 4 points are G(-1,3), H(2,1), F(3,0), and E(0,-2).
I'm trying to solve for the distance of diagonals GE and HF.
I added GE together by using the distance formula:
√(0- -1)^2+(-2-3)^2
√(0+1)^2+(-2-3)^2
√(1)^2+(-5)^2
√1+25
GE = √26
HF(2,1)(3,0)
√3-2)^2+(0-1)^2
√(-5)^2+(-1)^2
√25+1
√26
HF = √26
For both I got 26. However this is wrong. Can someone provide an explanation what I'm doing wrong? I'm also suppose to solve for their slopes but need to get the right distance for both. If anyone can provide an help that would be great.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The 4 points are G(-1,3), H(2,1), F(3,0), and E(0,-2).
I'm trying to solve for the distance of diagonals GE and HF.
I added GE together by using the distance formula:
√(0- -1)^2+(-2-3)^2
√(0+1)^2+(-2-3)^2
√(1)^2+(-5)^2
√1+25
GE = √26
HF(2,1)(3,0)
√3-2)^2+(0-1)^2
√(-5)^2+(-1)^2 ********* 3-2 = 1, not 5
√25+1
√26
HF = √26
For both I got 26. However this is wrong. Can someone provide an explanation what I'm doing wrong? I'm also suppose to solve for their slopes but need to get the right distance for both. If anyone can provide an help that would be great.
===================
HF = sqrt(1^2 + 1^2) = sqrt(2)
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