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

 Mar 5, 2020
 #1
avatar+111435 
+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)

 

 

cool cool cool

 Mar 5, 2020
edited by CPhill  Mar 5, 2020
 #2
avatar+25478 
+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 \(\mathbf{(4,~3)}\)

 

laugh

 Mar 5, 2020

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