coordinate systemsgeometrylinear algebra
The given vertices are (0,0) (5,-1) (-2,3).
My approach::
I assumed the orthocentre of the triangle as (a,b).
Now,
I used the fact that the line through each vertex and orthocentre is perpendicular to the opposite side.
So,
m (AH).m (BC) = -1
m (BH).m (AC) = -1
From this, I got two equations and I solved them to get the orthocentre.
However,
THIS METHOD IS VERY LENGTHY.
Can someone please suggests me easier or shorter method to find it?
Thanks.
(I assume that you meant "the orthocenter of the triangle is the origin.")
Let $A$ be $(5,-1)$, $B$ be $(-2,3)$, $O$ be the origin, and your desired third vertex be $C$.
Find the line through $O$ perpendicular to $\overline {AB}$: this will be the altitude of the triangle through $C$. (I get $-7x+4y=0$.) Then find the line through $A$ perpendicular to $\overline {BO}$: this will define the side of the triangle opposite to $B$, so this line also goes through $C$. (I get $-2x+3y=13$.)
The intersection of those two lines is $C$, your third vertex of the triangle. (I find the point to be $(-4,-7)$.)
You can check your answer by seeing that the line through $A$ and $O$ is perpendicular to the line through $B$ and $C$.
I'm sure you know the easy way to find perpendicular lines and to check that two lines are perpendicular.
The orthocenter of $ABC$ is the incenter of the orthic triangle.
If you check that $(10,15)$ is the incenter of the orthic triangle (or simply notice it is the only point lying inside the orthic triangle) it follows that $(5,10),(15,30),(50,-5)$ are the vertices of the original triangle, also since (by computing dot products) the given options form a orthocentric system.
Best Answer
I don't think there is a much better way to do that. One particular case, that is very convenient, is when the origin lies on the circumcenter, on that case because of the Euler's Line we get:
$$\text{orthocenter}=A'+B'+C'$$
But, in order to use that result you have to make a translation to bring the origin to the circumcenter and use that translation to find the new vertex $A',B',C'$.