The centroid of a triangle is $(2, 7)$ and two of its vertices are $(4, 8)$ and $(-2, 6)$. The third vertex is

In $\Delta ABC$,the coordinates of $B$ are $(0, 0)$,$AB = 2$,$\angle ABC = \frac{\pi}{3}$,and the coordinates of the midpoint of $BC$ are $(2, 0)$. Find the centroid of the triangle.

Let $ABC$ be an equilateral triangle with orthocenter at the origin and the side $BC$ on the line $x+2\sqrt{2}y=4$. If the coordinates of the vertex $A$ are $(\alpha, \beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta|$ is

The orthocenter of the triangle formed by the lines $4x - 7y + 10 = 0$,$x + y - 5 = 0$,and $7x + 4y - 15 = 0$ is:

The equation of the line joining the centroid with the orthocentre of the triangle formed by the points $(-2, 3), (2, -1), (4, 0)$ 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 . . . . . .