# What is the orthocenter of a triangle with corners at (4 ,1 ), (1 ,3 ), and (5 ,2 )?

Jul 26, 2018

The orthocenter of triangle is $\left(\frac{19}{5} , \frac{1}{5}\right)$

#### Explanation:

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

$A \left(4 , 1\right) , B \left(1 , 3\right) \mathmr{and} C \left(5 , 2\right)$

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 $\overline{A B} = \frac{1 - 3}{4 - 1} = - \frac{2}{3}$

$\overline{A B} \bot \overline{C N} \implies$slope of $\overline{C N} = \frac{3}{2}$ ,

$\overline{C N}$ passes through $C \left(5 , 2\right)$

$\therefore$The equn. of $\overline{C N}$ is $: y - 2 = \frac{3}{2} \left(x - 5\right)$

$\implies 2 y - 4 = 3 x - 15$

i.e. color(red)(3x-2y=11.....to (1)

Slope of $\overline{B C} = \frac{2 - 3}{5 - 1} = - \frac{1}{4}$

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

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

$\implies y - 1 = 4 x - 16$

i.e. color(red)(y=4x-15.....to (2)

Subst. $y = 4 x - 15$ into $\left(1\right)$ ,we get

$3 x - 2 \left(4 x - 15\right) = 11 \implies 3 x - 8 x + 30 = 11$

$- 5 x = - 19$

=>color(blue)( x=19/5

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

y=4(19/5)-15=>y=(76-75)/5=>color(blue)(y=1/5#

Hence, the orthocenter of triangle is $\left(\frac{19}{5} , \frac{1}{5}\right) = \left(3.8 , 0.2\right)$