Find the orthocenter of a Triangle JKL with vertices J(2,1), K(9,1), and L(4,6).

Guest Mar 5, 2020

#1**+3 **

The orthocenter is the intersection point of a triangle's altitudes.....we have....

Slope of line containing side JL = [6-1]/ [ 4 - 2] = 5/2

Equation of perpendicular line passing through K :

y = (-2/5)(x - 9) + 1

y = (-2/5)x + 18/5 + 1

y = (-2/5)x + 23/5 (1)

Slope of line containing side KL = [ 6 - 1] / [4 - 9] = 5/-5 = -1

Equation of a perpendicular line passing through J

y = 1 ( x - 2) + 1

y = x - 1 (2)

Equate (1) and (2) to find the x coordinate of the orthocenter

(-2/5)x + 23/5 = x - 1 rearrange as

23/5 + 1 = ( 2/5 + 1)x

28/5 = (7/5)x

28 = 7x

4 = x

And

y = 4 -1 = 3

So....the orthocenter is (4,3)

CPhill Mar 5, 2020

#2**+2 **

**Find the orthocenter of a Triangle JKL with vertices J(2,1), K(9,1), and L(4,6).**

\(\begin{array}{|rcll|} \hline x_{\text{Ortho}} &=& x_L \quad &| \quad x_L = 4 \\ \mathbf{x_{\text{Ortho}}} &=& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \tan{45^\circ} &=& \dfrac{2}{y} \quad &| \quad \tan{45^\circ}=1 \\ 1 &=& \dfrac{2}{y} \\ y &=& 2 \\\\ y_{\text{Ortho}} &=& y +y_J \quad &| \quad y_J = 1 \\ y_{\text{Ortho}} &=& 2 +1 \\ \mathbf{y_{\text{Ortho}}} &=& \mathbf{3} \\ \hline \end{array}\)

The orthocenter is \(\mathbf{(4,~3)}\)

heureka Mar 5, 2020