What will be the other vertex of the triangle? Two vertices of a triangles are $(5,-1)$ and  $(-2,3)$. If the orthocenter of the triangle is the origin, what is the other vertex ? 
My approach was that since the three vertices and the orthocenter form an orthic system, I need to find the orthocenter of the triangle with the three given points. Thus make two equations and solve... but it is becoming too cumbersome,is there any better method ?
 A: (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.
A: Hint:
If $A=(-2,3)$ and $B=(5-1)$, the two heights from $A$ and $B$ passe thorough the origin, so you can easely find the equation of these two straight line. The sides of the triangle opposite to $A$ and $B$ are orthogonal to these lines and you can find their equations, whose intersection is the other vertex of the triangle.
Added after comments.
Since the height from  A passes from $(0,0)$, its slope is $m=\dfrac{-3}{2}$, so the opposite side is the straight line from $B$ with slope $m'=\dfrac{2}{3}$ and his equation is  $y+1=\dfrac{2}{3}(x-5)$. 
And the same for the other side..... 
A: Let $C$ be the third vertex of the triangle. Let $ (x_A, y_A) $ and $ (x_B, y_B) $ be the coordinates of $ A $ and $ B $ respectively. The straight lines $ (HA) $ and $ (HB) $ have equations: 
$$ y_A x-x_Ay = 0,\qquad y_B x-x_By = 0,$$ 
hence the sides $ (AC) $, which is perpendicular to $ (HB) $, and $ (BC) $, perpendicular to $ (HA) $, have equations: $$ x_B x + y_By =x_B x_A + y_By_A\enspace \text{and}\quad x_A x + y_Ay =x_A x_B +y_A y_B$$ 
respectively. 
Gauß’s elimination results in the coordinates of the intersection $C$ of these sides: 
$$\begin{cases}
  x = \dfrac{(x_A x_B +y_A y_B)(y_B-y_A)}{x_A y_B - x_By_A}\\
  y = \dfrac{(x_A x_B +y_A y_B)(x_A-x_B)}{x_A y_B - x_By_A}
\end{cases} $$ 
Numerically, one gets $x_C=\color{red}{-7}$, $\,y_C=\color{red}{-4}$.
