Expert: Steve Holleran - 10/21/2007

Question
Two vertices of an isosceles triangle are (-5,4) and (3,8). The third vertex is on the x-axis. Find the possible coordinates of the 3rd vertex.

Answer
Hi Safwat,

Okay let's start by labeling A(-5,4) and B(3,8).

If the third vertex is on the x-axis, call it C(x,0).

If the triangle is to be isosceles, then AC = BC.
Using the distance formula,

AC = sqrt[(x+5)^2 + (0-4)^2] = sqrt[x^2 + 10x + 25 + 16}

  = sqrt[x^2 + 10x +41]

BC = sqrt[(x-3)^2 + (0-8)^2] = sqrt[x^2 -6x + 9 + 64]

  = sqrt[x^2 -6x + 73]

Since you want AC = BC

sqrt[x^2 + 10x + 41]  = sqrt[x^2 - 6x + 73]

squaring both sides,

    x^2 + 10x + 41 = x^2 - 6x + 73

The x^2's cancel, so

        10x + 41 = -6x + 73

        16x = 32  so x = 2.

So the third vertex is at (2, 0).

On the other hand, if the triangle is isosceles with AB = BC
then you have

AB = sqrt[(-5-3)^2 + (4-8)^2 ]

  = sqrt[64 + 16) = sqrt(80)

and

BC =  sqrt[(x-3)^2 + (0-8)^2] = sqrt[x^2 -6x + 9 + 64]

  = sqrt[x^2 -6x + 73]

Then sqrt[80] = sqrt[x^2 -6x +73]

so        80 = x^2 -6x  + 73

          0 = x^2 - 6x - 7

          0 = (x - 7)(x + 1)  so x =7 or x = -1

Then the third vertex would be (7, 0) or (-1, 0).

Hope this is what you needed
Steve