What is the orthocenter of a triangle with corners at (3 ,1 ), (4 ,5 ), and (2 ,7 )?

Jan 5, 2018

Orthocenter coordinates color(red)((-1/3), (-7/3)

Explanation: Slope of $B C = {m}_{b c} = \frac{{y}_{b} - {y}_{c}}{{x}_{b} - {x}_{c}} = \frac{5 - 7}{4 - 2} = - 1$

Slope of AD = m_(ad) = - (1/m_(bc) = - (1/-1) = 1#

Equation of AD is
$y - 1 = \left(1\right) \cdot \left(x - 3\right)$

$\textcolor{red}{- x + y = - 2}$ Eqn (1)

Slope of $A B = {m}_{A B} = \frac{{y}_{a} - {y}_{b}}{{x}_{a} - {x}_{b}} = \frac{1 - 5}{3 - 4} = 4$

Slope of $C F = {m}_{C F} = - \left(\frac{1}{m} _ \left(A B\right) = - \left(\frac{1}{4}\right)\right)$

Equation of CF is
$y - 7 = \left(4\right) \cdot \left(x - 2\right)$

$\textcolor{red}{- 4 x + y = - 1}$ Eqn (2)

Solving Eqns (1) & (2), we get the orthocenter $\textcolor{p u r p \le}{O}$ of the triangle

Solving the two equations,
$x = - \left(\frac{1}{3}\right) , y = - \left(\frac{7}{3}\right)$

Coordinates of orthocenter $\textcolor{p u r p \le}{O \left(- \frac{1}{3} , - \frac{7}{3}\right)}$