If two vertices of a triangle are $(5, -1)$ and $(-2, 3)$ and its orthocentre is at $(0, 0)$,then the third vertex is:

For a triangle formed by the vertices $(0,0)$,$(4,0)$,and $(3,4)$,the orthocenter is:

Let $A(6,8)$,$B(10 \cos \alpha, -10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$ be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ are its orthocenter and centroid respectively,then $(5a - 3h + 6k + 100 \sin 2\alpha)$ is equal to . . . . . .

What is the orthocenter of the triangle with vertices $(2, \frac{\sqrt{3}-1}{2})$,$(\frac{1}{2}, -\frac{1}{2})$,and $(2, -\frac{1}{2})$?

The base of an equilateral triangle with side $2a$ lies along the $y$-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

The orthocentre of the triangle formed by lines $x+y+1=0$,$x-y-1=0$,and $3x+4y+5=0$ is