What is the orthocenter of a triangle with corners at (2 ,8 ), (3 ,4 ), and (6 ,3 )#?

1 Answer

(-2/11, 16/11)

Explanation:

Let the vertices of triangle ABC be A(2, 8), B(3,4) & C(6, 3)

Now, the equation of altitude drawn from vertex A(2, 8) which is perpendicular to the opposite side BC

y-8=\frac{-1}{\frac{4-3}{3-6}}(x-2)

y-8=3(x-2)

3x-y=-2\ .........(1)

Similarly, the equation of altitude drawn from vertex B(3, 4) which is perpendicular to the opposite side AC

y-4=\frac{-1}{\frac{8-3}{2-6}}(x-3)

y-4=4/5(x-3)

4x-5y=-8\ .........(2)

Multiplying (1) by 5 & then subtracting from (2) we get

4x-5y-(15x-5y)=-8-(-10)

-11x=2

x=-2/11

\implies y=3x+2=3(-2/11)+2=16/11

The ortho-center is the point of intersection of altitudes drawn from vertices to the opposite sides of a triangle hence the orthocenter of given triangle is

(-2/11, 16/11)