Math Help - Find Orthocenter of Triangle

1. Find Orthocenter of Triangle

I need to find the orthocenter of a triangle with coordinates:

G(-2,5)
H(6,5)
J(4,-1)

AND...

A(4,-3)
B(8,5)
C(8,-8)

2. Let O(a,b) be the orthocenter of triangle GHJ. Since O is the orthocenter, OG is perpendicular to HJ and OH is perpendicular to GJ.

Thus (slope of OG) x (slope of HJ) = -1. Also (slope of OH) x (slope of GJ) = -1. Those two slope equations will you give you two simultaneous equations in a and b. Solve them and get O.

3. ok........

Originally Posted by Isomorphism
Let O(a,b) be the orthocenter of triangle GHJ. Since O is the orthocenter, OG is perpendicular to HJ and OH is perpendicular to GJ.

Thus (slope of OG) x (slope of HJ) = -1. Also (slope of OH) x (slope of GJ) = -1. Those two slope equations will you give you two simultaneous equations in a and b. Solve them and get O.
Thanks for the tips.

4. Hello, magentarita!

The orthocenter is the intersection of the altitudes.
Did you make a sketch ??

Find the orthocenter of a triangle with vertices:
G(-2,5), H(6,5), J(4,-1)
Code:
|
(-2,5)    |          (6,5)
A o  *  *  *  *  *  o B
*   |     :    *
* |     :   *
*     :  *
- - - - + * - : * - - - - - -
|   * :*
|     o C
|  (4,-1)
|

Recall that an altitude is perpendicular to the opposite side.

Since $AB$ is horizontal,
. . we see the altitude from $C$ is the vertical line $x = 4$

The slope of $AC$ is: . $m_1 \:=\:\frac{\text{-}1-5}{4-(\text{-}2)} \:=\:-1$

Hence, the altitude from $B$ has slope: . $m_2 \:=\:1$

The equation of altitude from $B(6,5)$ with slope $1$
. . $y - 5 \:=\:1(x-6) \quad\Rightarrow\quad y \:=\:x-1$

The intersection of this altitude with the vertical line $x = 4$

. . is: . $\begin{Bmatrix}y \:=\:x-1 \\ x \:=\:4\end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x \:=\:4\\y\:=\:3 \end{Bmatrix}$

Therefore, the orthocenter is $(4,3)$

5. great...

Originally Posted by Soroban
Hello, magentarita!

The orthocenter is the intersection of the altitudes.
Did you make a sketch ??

Code:
|
(-2,5)    |          (6,5)
A o  *  *  *  *  *  o B
*   |     :    *
* |     :   *
*     :  *
- - - - + * - : * - - - - - -
|   * :*
|     o C
|  (4,-1)
|
Recall that an altitude is perpendicular to the opposite side.

Since $AB$ is horizontal,
. . we see the altitude from $C$ is the vertical line $x = 4$

The slope of $AC$ is: . $m_1 \:=\:\frac{\text{-}1-5}{4-(\text{-}2)} \:=\:-1$

Hence, the altitude from $B$ has slope: . $m_2 \:=\:1$

The equation of altitude from $B(6,5)$ with slope $1$
. . $y - 5 \:=\:1(x-6) \quad\Rightarrow\quad y \:=\:x-1$

The intersection of this altitude with the vertical line $x = 4$

. . is: . $\begin{Bmatrix}y \:=\:x-1 \\ x \:=\:4\end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x \:=\:4\\y\:=\:3 \end{Bmatrix}$

Therefore, the orthocenter is $(4,3)$
Another great job. Happy New Year, Soroban.