What is the orthocenter of a triangle with corners at (3 ,2 ), (4 ,5 ), and (2 ,7 )?

Nov 4, 2017

Orthocenter of the triangle is at $\left(5.5 , 6.5\right)$

Explanation:

Orthocenter is the point where the three "altitudes" of a triangle meet. An "altitude" is a line that goes through a vertex (corner point) and is at right angles to the opposite side.

$A = \left(3 , 2\right) , B \left(4 , 5\right) , C \left(2 , 7\right)$ . Let $A D$ be the altitude from $A$ on $B C$ and $C F$ be the altitude from $C$ on $A B$ they meet at point $O$ , the orthocenter.

Slope of $B C$ is ${m}_{1} = \frac{7 - 5}{2 - 4} = - 1$
Slope of perpendicular $A D$ is ${m}_{2} = 1 \left({m}_{1} \cdot {m}_{2} = - 1\right)$

Equation of line $A D$ passing through $A \left(3 , 2\right)$ is $y - 2 = 1 \left(x - 3\right)$ or
$y - 2 = x - 3 \mathmr{and} x - y = 1 \left(1\right)$

Slope of $A B$ is ${m}_{1} = \frac{5 - 2}{4 - 3} = 3$
Slope of perpendicular $C F$ is ${m}_{2} = - \frac{1}{3} \left({m}_{1} \cdot {m}_{2} = - 1\right)$

Equation of line $C F$ passing through $C \left(2 , 7\right)$ is $y - 7 = - \frac{1}{3} \left(x - 2\right)$ or
$y - 7 = - \frac{1}{3} x + \frac{2}{3} \mathmr{and} \frac{1}{3} x + y = 7 + \frac{2}{3} \mathmr{and} \frac{1}{3} x + y = \frac{23}{3}$ or
$x + 3 y = 23 \left(2\right)$

Solving equation(1) and (2) we get their intersection point , which is the orthocenter.

x-y=1 (1) ; x+3y=23(2) Subtracting (1) from (2) we get,
4y=22 :. y=5.5 ; x = y+1=6.5#

Orthocenter of the triangle is at $\left(5.5 , 6.5\right)$ [Ans]