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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 geth=-4,k=-7\,

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

Main Page:Geometry:Straight Lines-I