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

$\left(- \frac{2}{11} , \frac{16}{11}\right)$

#### Explanation:

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

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

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

$y - 8 = 3 \left(x - 2\right)$

$3 x - y = - 2 \setminus \ldots \ldots \ldots \left(1\right)$

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

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

$y - 4 = \frac{4}{5} \left(x - 3\right)$

$4 x - 5 y = - 8 \setminus \ldots \ldots \ldots \left(2\right)$

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

$4 x - 5 y - \left(15 x - 5 y\right) = - 8 - \left(- 10\right)$

$- 11 x = 2$

$x = - \frac{2}{11}$

$\setminus \implies y = 3 x + 2 = 3 \left(- \frac{2}{11}\right) + 2 = \frac{16}{11}$

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

$\left(- \frac{2}{11} , \frac{16}{11}\right)$