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

Sep 28, 2017

$\left(- 6. \overline{3} , - 1. \overline{3}\right)$

#### Explanation:

$L e t$ $A = \left(3 , 1\right)$
$L e t$ $B = \left(1 , 6\right)$
$L e t$ $C = \left(2 , 2\right)$

Equation for altitude through A:
$x \left({x}_{3} - {x}_{2}\right) + y \left({y}_{3} - {y}_{2}\right) = {x}_{1} \left({x}_{3} - {x}_{2}\right) + y 1 \left({y}_{3} - {y}_{2}\right)$
$\implies x \left(2 - 1\right) + y \left(2 - 6\right) = \left(3\right) \left(2 - 1\right) + \left(1\right) \left(2 - 6\right)$
$\implies x - 4 y = 3 - 4$
$\implies \textcolor{red}{x - 4 y + 1 = 0}$-----(1)

Equation for altitude through B:
$x \left({x}_{1} - {x}_{3}\right) + y \left({y}_{1} - {y}_{3}\right) = {x}_{2} \left({x}_{1} - {x}_{3}\right) + y 2 \left({y}_{1} - {y}_{3}\right)$
$\implies x \left(3 - 2\right) + y \left(1 - 2\right) = \left(1\right) \left(3 - 2\right) + \left(6\right) \left(1 - 2\right)$
$\implies x - y = 1 - 6$
=>color(blue)(x-y+5=0-----(2)

Equating (1) & (2):
color(red)(x-y+5)=color(blue)(x-4y+1
$\implies - y + 4 = 1 - 5$
=>color(orange)(y=-4/3-----(3)

Plugging (3) in (2):
$\textcolor{b l u e}{x - 4} \textcolor{\mathmr{and} a n \ge}{\left(- \frac{4}{3}\right)} \textcolor{b l u e}{+ 1} = 0$
=>color(violet)(x=-19/3#

The orthocenter is at $\left(- \frac{19}{3} , - \frac{4}{3}\right)$ OR $\left(- 6.333 \ldots , - 1.333 \ldots\right)$
which is actually outside the $\triangle$ because the $\triangle$ is an obtuse $\triangle$. Click here to find more.