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?

Algebra ->  Points-lines-and-rays -> 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?      Log On


   

Question 1009436: 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?
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
Those two points could serve as segment for the base; understanding Distance Formula, and area formula of a triangle assumed, the length of that base for those two given points is sqrt%2817%29. The altitude of the triangle would be therefore h=%284%2F17%29sqrt%2817%29.

Consider two line equations.
The base is on y=x%2F4%2B7%2F2 and another line to contain the unknown triangle's vertex is some y=x%2F4%2Bb, and you do not yet know b.

You want the distance between the two lines to be %284%2F17%29sqrt%2817%29. Putting this as an equation, .

Understand that these are two parallel lines, and once you find b, ANY point on the line of the then found & solved b, will contain an acceptable "third" vertex, because any of them will be the h distance from the chosen base segment.


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(Strategy as as much work as shown, done on paper - not finished beyond how was described here.)

Note that b=%28119%2B-+8sqrt%2817%29%29%2F34.