The centroid of a triangle is $(2, 3)$ and two of its vertices are $(5, 6)$ and $(-1, 4)$. Find the third vertex of the triangle.

The coordinates of the orthocenter of the triangle whose sides are represented by the lines $L_1: 4x - 7y + 10 = 0$, $L_2: x + y - 5 = 0$ and $L_3: 7x + 4y - 15 = 0$ are...

If $(\alpha, \beta)$ is the orthocentre of the triangle $ABC$ with vertices $A(3, -7)$,$B(-1, 2)$,and $C(4, 5)$,then $9\alpha - 6\beta + 60$ is equal to:

The orthocentre of the triangle with vertices $\left( 2, \frac{\sqrt{3} - 1}{2} \right)$,$\left( \frac{1}{2}, -\frac{1}{2} \right)$,and $\left( 2, -\frac{1}{2} \right)$ is:

$(-2, -1)$ and $(2, 5)$ are two vertices of a triangle and $\left(2, \frac{5}{3}\right)$ is its orthocenter. If $(m, n)$ is the third vertex of that triangle,then $m+n=$

In a triangle $ABC$,the coordinates of $A$ are $(1, 2)$ and the equations of the medians through $B$ and $C$ are $x + y = 5$ and $x = 4$ respectively. Then the area of $\Delta ABC$ (in sq. units) is