# Orthocenter...what is it with this triangle

• Mar 7th 2009, 05:27 PM
squat4speed
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
• Mar 7th 2009, 05:46 PM
lanzailan
Quote:

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...
• Mar 7th 2009, 05:57 PM
squat4speed
how do i find coordinates of point of intersection
• Mar 7th 2009, 06:04 PM
squat4speed
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.
• Mar 7th 2009, 06:11 PM
lanzailan
Quote:

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).
• Mar 7th 2009, 06:17 PM
Soroban
Hello, squat4speed!

Quote:

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

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

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

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

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

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

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

• Mar 7th 2009, 06:25 PM
lanzailan
if you draw on the graph paper you will get (1.6,1)