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

Oct 8, 2016

The orthocenter is at the point $\left(\frac{4}{3} , - \frac{17}{3}\right)$

Explanation:

Let's begin by writing the equation of the line that goes through point $\left(3 , 1\right)$ and perpendicular to the line going through points $\left(1 , 3\right)$ and $\left(5 , 2\right)$.

The slope, m, of the line going through points $\left(1 , 3\right)$ and $\left(5 , 2\right)$ is:

$m = \frac{3 - 2}{1 - 5} = - \frac{1}{4}$

The slope, n, of any line perpendicular is

$n = - \frac{1}{m} = 4$

Use point slope form of the equation of a line to obtain the equation of a line that we desire:

$y - {y}_{1} = n \left(x - {x}_{1}\right)$

$y - 1 = 4 \left(x - 3\right)$

$y - 1 = 4 x - 12$

$y = 4 x - 11$

Write the equation of a line that goes through point $\left(5 , 2\right)$ perpendicular to the line through the points $\left(3 , 1\right)$ and $\left(1 , 3\right)$:

The slope, m, of the line that goes through the points $\left(3 , 1\right)$ and $\left(1 , 3\right)$ is:

$m = \frac{3 - 1}{1 - 3} = \frac{2}{-} 2 = - 1$

The slope, n, of any line perpendicular is:

$n = - \frac{1}{m} = 1$

Again, Use the point slope form for the point $\left(5 , 2\right)$:

$y - {y}_{1} = n \left(x - {x}_{1}\right)$

$y - 2 = 1 \left(x - 5\right)$

$y = x - 7$

The orthocenter is at the intersection of these two lines:

$y = 4 x - 11$
$y = x - 7$

$x - 7 = 4 x - 11$

$4 = 3 x$

$x = \frac{4}{3}$

$y = \frac{4}{3} - 7$

$y = - \frac{17}{3}$

The orthocenter is at the point $\left(\frac{4}{3} , - \frac{17}{3}\right)$