# Math Help - Orthocenter...what is it with this triangle

1. ## Orthocenter...what is it with this triangle

Whats the orthocenter of a triangle with coordinates R (-1 , -4) , P (-1 , 5) , Q (7 , 2).

Please show how you did it as simple as possible

Thanks

2. Originally Posted by squat4speed
Whats the orthocenter of a triangle with coordinates R (-1 , -4) , P (-1 , 5) , Q (7 , 2).

Please show how you did it as simple as possible

Thanks
i think you need to find the midpoint of the 2 sides of the triangle. from these new points you will come up with 2 equation (linear equation) and find the coordinates of the point of intersection. that would be your orthocenter..
i hope this is correct...

3. how do i find coordinates of point of intersection

4. Alright I got a head ache..and im confusing myself somehow, can someone please give me a step by step show of how you find the coordinates of the orthocenter.

5. Originally Posted by squat4speed
how do i find coordinates of point of intersection
when you got 2 equation what you need to do is solving using simultaneous or substitution method.

ok i got midpoint of RQ=(3,-1) <<---let this point be point A
and PR=(-1,1/2)<<-------- this point be point B

now find the equation of line that passes through point B and point Q (first equation)
and equation of line that passes through A and P(second equation).

6. Hello, squat4speed!

What's the orthocenter of a triangle with coordinates: $R (-1 , -4) ,\:P (-1 , 5) ,\: Q (7 , 2).$

The orthocenter is the intersection of the three altitudes of the triangle.

Find the equation of two of the altitudes and locate their intersection.

If you make a sketch, the problem is much simpler.
Code:
          P   |
(-1,5)*   |
* * |
*   *
*   | *
*   |   *
*   |     *
- - * - + - - - *Q
*   |     * (7,2)
*   |   *
------*---+-*----------
*   *
* * |
(-1,-4)*   |
R   |
|

Side $PR$ is vertical.
Hence, the altitude from $Q$ is the horizontal line, $y = 2.$ .[1]

Side QR has slope: . $m \:=\:\frac{2-(\text{-}4)}{7-(\text{-}1)} \:=\:\frac{3}{4}$

Hence, the altitude from $P(\text{-}1,5)$ has slope $\text{-}\tfrac{4}{3}$
Its equation is: . $y - 5 \:=\:\text{-}\tfrac{4}{3}(x+1) \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{4}{3}x + \tfrac{11}{3}$ .[2]

The intersection of [1] and [2]: . $2 \:=\:\text{-}\tfrac{4}{3}x + \tfrac{11}{3} \quad\Rightarrow\quad \tfrac{4}{3}x \:=\:\tfrac{5}{3}$

Therefore: . $x \:=\:\tfrac{5}{4},\;y \;=\:2$

The orthocenter is: . $\left(\tfrac{5}{4},\:2\right)$

7. if you draw on the graph paper you will get (1.6,1)