Geo2.45

Two vertices of a triangle are $(5,-1),(-2,3)\,$ . If the orthocenter of the triangle is the origin,find the third vertex.

Let $B(5,-1),C(-2,3)\,$ be the given vertices and $(h,k)\,$ be the third vertex.

Let AD and BE be the altitudes meeting at the orthocenter O(0,0) of the triangle.

Slope of $BC=\frac{3+1}{-2-5}=\frac{-4}{7},AD=AO=\frac{k-0}{h-0}=\frac{k}{h}\,$

AD is perpendicular to BC,therefore $\frac{k}{h}\cdot\frac{-4}{7}=-1,7h=4k\,$ let this be equation 1.

Slope of $AB=\frac{k+1}{h-5},CF=CO=\frac{3-0}{-2-0}=\frac{-3}{2}\,$

Similar to the above, $\frac{k+1}{h-5}\cdot\frac{-3}{2}=-1,2h-3k-13=0\,$ .Let this be equation 2.

Solving the two equations,we get $h=-4,k=-7\,$

Therefore the third vertex is $(-4,-7)\,$