SOLUTION: Two vertices of a triangle are (2,4) and (-2,3) and the area is 2 square units, the locus of the third vertex is? What should be the answer here?

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Question 1031169: Two vertices of a triangle are (2,4) and (-2,3) and the area is 2 square units, the locus of the third vertex is?
What should be the answer here?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Let distance between those two points be the base for the triangle.

sqrt%28%282-%28-2%29%29%5E2%2B%284-3%29%5E2%29
sqrt%2816%2B1%29
sqrt%2817%29

An altitude could intersect the base ANYWHERE and you want to be sure that the altitude length will give area of 2 square units. Let a be the altitude length.
%281%2F2%29sqrt%2817%29%2Aa=2
a%2Asqrt%2817%29=4
a=4%2Fsqrt%2817%29

THE PROBLEM HAS INFINITELY MANY SOLUTIONS. The altitude must be perpendicular to the base and be 4%2Fsqrt%2817%29.

BASE
y-3=%28%284-3%29%2F%282-%28-2%29%29%29%28x-%28-2%29%29
y-3=%281%2F4%29%28x%2B4%29
and perpendicular line containing the altitude is , if you choose one endpoint on the altitude to be (2,4), y-4=-4%28x-2%29
y=-4x%2B8%2B4
y=-4x%2B12------line containing the altitude at base endpoint (2,4).

An unknown point (x,-4x+12) must be distance sqrt%2817%29 from point (2,4).
-
sqrt%28%28x-2%29%5E2%2B%28-4x%2B12-4%29%5E2%29=sqrt%2817%29
sqrt%28%28x-2%29%5E2%2B%28-4%2B8%29%5E2%29=sqrt%2817%29
%28x-2%29%5E2%2B%288-4x%29%5E2=17
x%5E2-4x%2B4%2B64-64x%2B16x%5E2=17
17x%5E2-68x%2B68=17
17x%5E2-68x%2B57=0

x=%2868%2B-+sqrt%2868%5E2-4%2A17%2A57%29%29%2F%282%2A17%29
x=%2868%2B-+sqrt%28748%29%29%2F%282%2A17%29
x=%2868%2B-+2sqrt%28187%29%29%2F%282%2A17%29
highlight%28x=34%2B-+sqrt%28187%29%29%2F17%29-----Use this the find the corresponding y coordinate values.
x=34%2B-+sqrt%2811%2A17%29%29%2F17------not simplifiable further