Question 1058558: The vertices of the base of an isosceles *triangle* are
(1,2)and R (4,-1). Find the ordinate of the third vertex
if its abscissa is 6.
Found 3 solutions by solve_for_x, rothauserc, Edwin McCravy:
Answer by solve_for_x(190)   (Show Source):
You can put this solution on YOUR website! Since the line connecting (1, 2) and (4, -1) is the base of the isosceles triangle, the vertex
must be along the line x = 6, with coordinates (6, y).
Since the triangle is isosceles, the distance from (1, 2) to (6, y) must be equal to the distance from (4, -1)
to (6, y).
To make things simpler, and avoid using radicals, the square of the distance between the points will
be compared:
Between (1, 2) and (6, y):
d^2 = (6 - 1)^2 + (y - 2)^2 = 25 + (y - 2)^2
Between (4, -1) and (6, y):
d^2 = (6 - 4)^2 + (y + 1)^2 = 4 + (y + 1)^2
Equating the two squares gives:
25 + (y - 2)^2 = 4 + (y + 1)^2
25 + y^2 - 4y + 4 = 4 + y^2 + 2y + 1
Subtracting y^2 from both sides leaves:
25 - 4y + 4 = 4 + 2y + 1
29 - 4y = 5 + 2y
2y + 4y = 29 - 5
6y = 24
y = 24/6
y = 4
Solution: The ordinate of the third vertex is 4.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! We can use the distance formula for any two points in the Cartesian plane
:
we have three points (1,2), (4,-1), (6,y)
:
since the triangle is isosceles, we know that
:
square root( (6-1)^2 + (y-2)^2) ) = square root( (6-4)^2 + (y+1)^2 )
:
square both sides of the =
:
25 + y^2 -4x + 4 = 4 + y^2 +2x + 1
:
6y = -24
:
y = -4
:
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The ordinate of the third point is -4
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Answer by Edwin McCravy(20055)  (Show Source):
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