# 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 $(-4,-7)\,$