Expert: Josh - 10/28/2007

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 third vertex.

THIS IS NOT A HOMEWORK QUESTION!!!
IT CAME ON A TEST

Answer
Best we to see this is to draw a diagram. Draw the x,y axes on a piece of paper. Mark the two coordinates with a cross. The third vertex (X,Y) is constrained by the condition that it must be equidistant from these two points. This is needed to form an isosceles triangle. (Recall that two of the sides must have equal length)

We can guess the position of (X,Y) graphically. Or, we can solve this analytically.

Since the distance between (-5,4) and (X,Y) equals distance between (3,8) and (X,Y). Using the distance formula*,
sqrt((X-(-5))^2+(Y-4)^2) = sqrt((X-3)^2+(Y-8)^2)...squaring both sides
(X-(-5))^2+(Y-4)^2 = (X-3)^2+(Y-8)^2
X^2+10X+25+Y^2-8Y+16 = X^2-6X+9+Y^2-16Y+64 ...canceling common terms
10X-8Y+41 = -6X-16Y+73
16X+8Y=32
simplifying,
Y=-2X+4

You can check the working again to make sure that we haven't made any mistake.

Interpretation: Possible points for the third vertex must lie on the straight line described by the equation above. You can draw a sketch to see if this makes sense. Obviously impossible for me to do here.

Hope it's all good:)

* Note: The distance between two points (x1,y1) and (x2,y2) is given by square_root_of[(x2-x1)^2+(y2-y1)^2]