What is the orthocenter of a triangle with corners at (3 ,3 ), (2 ,4 ), and (7 ,9 )?

Jun 17, 2018

The orthocentre of $\triangle A B C$ is $B \left(2 , 4\right)$

Explanation:

We know$\text{ the "color(blue)"Distance Formula} :$

$\text{The distance between two points}$ $P \left({x}_{1} , {y}_{1}\right) \mathmr{and} Q \left({x}_{2} , {y}_{2}\right)$ is:

color(red)(d(P,Q)=PQ=sqrt((x_1-x_2)^2+(y_1-y_2)^2)...to(1)

Let , $\triangle A B C$ ,be the triangle with corners at

$A \left(3 , 3\right) , B \left(2 , 4\right) \mathmr{and} C \left(7 , 9\right)$ .

We take, $A B = c , B C = a \mathmr{and} C A = b$

So, using color(red)((1) we get

${c}^{2} = {\left(3 - 2\right)}^{2} + {\left(3 - 4\right)}^{2} = 1 + 1 = 2$

${a}^{2} = {\left(2 - 7\right)}^{2} + {\left(4 - 9\right)}^{2} = 25 + 25 = 50$

${b}^{2} = {\left(7 - 3\right)}^{2} + {\left(9 - 3\right)}^{2} = 16 + 36 = 52$

It is clear that, ${c}^{2} + {a}^{2} = 2 + 50 = 52 = {b}^{2}$

 i.e. color(red)(b^2=c^2+a^2=>m angle B=pi/2#

Hence, $\overline{A C}$ is the hypotenuse.

$\therefore \triangle A B C$ is the right angled triangle.

$\therefore$The orthocenter coindes with $B$

Hence, the orthocentre of $\triangle A B C$ is $B \left(2 , 4\right)$