The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is

Two vertices of a triangle $ABC$ are $A(3, -1)$ and $B(-2, 3)$,and its orthocentre is $P(1, 1)$. If the coordinates of the point $C$ are $(\alpha, \beta)$ and the centre of the circle circumscribing the triangle $PAB$ is $(h, k)$,then the value of $(\alpha + \beta) + 2(h + k)$ equals :

Let $O(0,0), P(3,4), Q(6,0)$ be the vertices of the triangle $OPQ$. The point $R$ inside the triangle $OPQ$ is such that the triangles $OPR, PQR, OQR$ are of equal area. The coordinates of $R$ are

If the orthocentre of the triangle formed by the lines $2x+3y-1=0$,$x+2y-1=0$,and $ax+by-1=0$ is the centroid of another triangle,whose circumcentre and orthocentre are $(3,4)$ and $(-6,-8)$ respectively,then the value of $|a-b|$ 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 . . . . . .

If $\left( \frac{3}{2}, 0 \right)$,$\left( \frac{3}{2}, 6 \right)$,and $(-1, 6)$ are the midpoints of the sides of a triangle,find the centroid of the triangle.