# What is the orthocenter of a triangle with corners at (1, 3), (6, 2), and (5, 4)?

Apr 22, 2018

$\left(x , y\right) = \left(\frac{47}{9} , \frac{46}{9}\right)$

#### Explanation:

Let: A(1, 3), B(6, 2) and C(5, 4) be the vertices of triangle ABC:

Slope of a line through points: $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$:
$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

Slope of AB:
$= \frac{2 - 3}{6 - 1} = - \frac{1}{5}$
Slope of perpendicular line is 5.
Equation of the altitude from C to AB:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$ =>$m = 5 , C \left(5 , 4\right)$:
$y - 4 = 5 \left(x - 5\right)$
$y = 5 x - 21$

Slope of BC:
$= \frac{4 - 2}{5 - 6} = - 2$
Slope of perpendicular line is 1/2.
Equation of the altitude from A to BC:
$y - 3 = \frac{1}{2} \left(x - 1\right)$
$y = \left(\frac{1}{2}\right) x + \frac{5}{2}$

The intersection of the altitudes equating y's:
$5 x - 21 = \left(\frac{1}{2}\right) x + \frac{5}{2}$
$10 x - 42 = x + 5$
$9 x = 47$
$x = \frac{47}{9}$

$y = 5 \cdot \frac{47}{9} - 21$
$y = \frac{46}{9}$

Thus the Orthocenter is at $\left(x , y\right) = \left(\frac{47}{9} , \frac{46}{9}\right)$

To check the answer you can find the equation of altitude from B to AC and find the intersection of that with one of the other altitudes.