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Question 198664: the vertices of the base of an isosceles triangle are (1,2) & (4,-1). find the ordinate of the third vertex if its abcissa is 6
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! the vertices of the base of an isosceles triangle are (1,2) & (4,-1). find the ordinate of the third vertex if its abcissa is 6
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The 3rd vertex has to be equidistant from the 2 given points, and it's on the line x = 6. The point will be (6,y).
The distance from (1,2) is sqrt of ((6-1)^2 + (y-2)^2)
The distance from (4,-1) is sqrt of ((6-4)^2 + (y+1)^2)
The distances are equal, so the squares are equal.
5^2 + (y-2)^2 = 2^2 + (y+1)^2
y^2 - 4y + 29 = y^2 + 2y + 5
-6y = -24
y = 4
Vertex at (6,4)
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