Question 1045199: Two vertices of an equilateral triangle are (10,-4) and (0,6). Find the third vertex
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Two vertices of an equilateral triangle are (10,-4) and (0,6). Find the third vertex
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Label the points A(10,-4) and B(0,6)
Find the length of AB and the midpoint, label it M
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Length of AB = sqrt(diffy^2 + diffx^2) = sqrt(100+100) = 10sqrt(2)
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Find the average of x & y separately.
(10+0)/2 = 5
(-4+6)/2 = 1
M(5,1)
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Find the equation of the perpendicular bisector.
Slope of AB = 10/-10 = -1
Slope m of the perpendicular bisector = +1
--> y-1 = 1*(x-5)
y = x-4
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There are 2 vertices, C & D, both are on the line y = x-4
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The altitude of the triangle is 10.
Find the 2 points on y = x-4 10 units distance from M(5,1)
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d^2 = diffx^2 + diffy^2
100 = (x-5)^2 + (y-1)^2
Sub for y
100 = (x-5)^2 + (x-5)^2
(x-5)^2 = 50
x-5 = ħsqrt(50)
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x = 5 + sqrt(50), y = 1 + sqrt(50) --> Vertex C
x = 5 - sqrt(50), y = 1 - sqrt(50) --> Vertex D
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