# What is the orthocenter of a triangle with corners at (1 ,3 ), (6 ,9 ), and (2 ,4 )?

Aug 5, 2018

The orthocenter of triangle is $\left(56 , - 41\right)$

#### Explanation:

Let $\triangle A B C \text{ be the triangle with corners at}$

$A \left(1 , 3\right) , B \left(6 , 9\right) \mathmr{and} C \left(2 , 4\right)$

Let $\overline{A L} , \overline{B M} \mathmr{and} \overline{C N}$ be the altitudes of sides $\overline{B C} , \overline{A C} \mathmr{and} \overline{A B}$ respectively.

Let $\left(x , y\right)$ be the intersection of three altitudes .

Slope of $\overline{A B} = \frac{3 - 9}{1 - 6} = \frac{6}{5}$

$\overline{A B} \bot \overline{C N} \implies$slope of $\overline{C N} = - \frac{5}{6}$ ,

$\overline{C N}$ passes through $C \left(2 , 4\right)$

$\therefore$The equn. of $\overline{C N}$ is $: y - 4 = - \frac{5}{6} \left(x - 2\right)$

$6 y - 24 = - 5 x + 10$

i.e. color(red)(5x+6y=34.....to (1)

Slope of $\overline{B C} = \frac{9 - 4}{6 - 2} = \frac{5}{4}$

$\overline{A L} \bot \overline{B C} \implies$slope of $\overline{A L} = - \frac{4}{5}$ , $\overline{A L}$ passes through $A \left(1 , 3\right)$

$\therefore$The equn. of $\overline{A L}$ is $: y - 3 = - \frac{4}{5} \left(x - 1\right)$

$\implies 5 y - 15 = - 4 x + 4$

=>color(red)(4x+5y=19.....to (2)

Taking $e q n . \left(1\right) \times 5 - e q n . \left(2\right) \times \left(- 6\right)$ and adding

$\textcolor{w h i t e}{\ldots .} 25 x + 30 y = 170$
ul(-24x-30y=-114
$\therefore 1 x - 0 = 56$

=>color(blue)( x=56

From equn.$\left(2\right)$ we get

4(56)+5y=19=>5y=19-224=-205=>color(blue)(y=-41#

Hence, the orthocenter of triangle is $\left(56 , - 41\right)$