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Question 1094532: Three points D(-5,6), E(2,-1), and F(x,0) are given. Find the value of X if DF=EF.
Found 3 solutions by mananth, ikleyn, greenestamps:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! find the distances DF and EF using the distance formula:
Distance formula: d = sqrt((x₂ - x₁)^2 + (y₂ - y₁)^2)
DF = sqrt((-5 - x)^2 + (6 - 0)^2)
EF = sqrt((2 - x)^2 + (-1 - 0)^2)
sqrt((-5 - x)^2 + (6 - 0)^2) = √((2 - x)^2 + (-1 - 0)^2)
Squaring both sides of the equation to eliminate the square roots:
((-5 - x)^2 + (6 - 0)^2) = ((2 - x)^2 + (-1 - 0)^2)
simplify
25+10x+x^2 +36=4-4x+x^2+1
14x=-56
x=-56/14
Answer by ikleyn(52810)  (Show Source):
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Here is an alternative solution method, using a bit of geometry along with the algebra.
If DF=EF, then F is on the perpendicular bisector of DE.
The midpoint of DE is (-1.5,2.5), and its slope is -1.
So F is on the line with slope 1 passing through (-1.5,2.5). The point-slope form of the equation of that line is
y-2.5 = 1(x+1.5)
We know F is (x,0), so y is 0. Substitute y=0 and solve to find x.
0-2.5 = x+1.5
-2.5-1.5 = x
x = -4
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