# Answer to Question #19984 in Geometry for robbie schwartz

Question #19984

Point A is on the line y+3x=2 and is equidistant from points B and C, both of which lie on the line y=1/3x+8. D is on the intersection of both lines. Prove that y+3x=2 bisects BC.

Expert's answer

AC=BC, that means, that

Sqrt((Ay-Cy)^2+(Ax-Cx)^2=Sqrt((By-Cy)^2+(Bx-Cx)^2), hence

Sqrt((2-3Ax-1/3Cx-8)^2+(Ax-Cx)^2=Sqrt((2-3Bx-1/3Cx-8)^2+(Bx-Cx)^2)

Hence, we see, that the only point, that is satisfying thisequation are points, that are equidistant from line intersection.

That means, line is bisecting BC

Sqrt((Ay-Cy)^2+(Ax-Cx)^2=Sqrt((By-Cy)^2+(Bx-Cx)^2), hence

Sqrt((2-3Ax-1/3Cx-8)^2+(Ax-Cx)^2=Sqrt((2-3Bx-1/3Cx-8)^2+(Bx-Cx)^2)

Hence, we see, that the only point, that is satisfying thisequation are points, that are equidistant from line intersection.

That means, line is bisecting BC

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