If the equation of the median through vertex $A(3, k)$ of $\triangle ABC$ with vertices $B(2, 1)$ and $C(-4, 5)$ is $x + 4y = p$,then $k = ?$ where $p$ and $k$ are constants.

If $P(6,1)$ is the orthocentre of the triangle whose vertices are $A(5,-2)$,$B(8,3)$,and $C(h, k)$,then the point $C$ lies on the circle:

Find the incenter of the triangle with vertices $(1, \sqrt{3})$,$(0, 0)$,and $(2, 0)$.

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 with the vertices $(2,2), (5,1), (4,4)$,then $\alpha+\beta=$

If in triangle $ABC$,$A \equiv (1, 10)$,circumcentre $\equiv \left( -\frac{1}{3}, \frac{2}{3} \right)$ and orthocentre $\equiv \left( \frac{11}{3}, \frac{4}{3} \right)$,then the coordinates of the mid-point of the side opposite to $A$ are: