SOLUTION: 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.

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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
:
*************************************
The ordinate of the third point is -4
*************************************
:

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Darn, I misread ordinate and abscissa!
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.

You have this:



and you want to find x.

Since the triangle is isosceles, we use the distance formula
to set the lengths of the two green sides equal to each other:



Square both sides. (Take away the square root symbols)



Simplify:







Get the squares on the left and the number on the right:



Factor the left as a difference of squares













So the abscissa (x-coordinate) of the vertex of the triangle
is 8, and so the vertex is the point (8,6).



Edwin
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