Two sides of a parallelogram are along the lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of one of the diagonals of the parallelogram is $11 \mathrm{x}+7 \mathrm{y}=9$, then other diagonal passes through the point:

Given three points $P, Q, R$ with $P(5, 3)$ and $R$ lies on the $x-$ axis. If equation of $RQ$ is $x - 2y = 2$ and $PQ$ is parallel to the $x-$ axis, then the centroid of $\Delta PQR$ lies on the line

Locus of the points which are at equal distance from $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$ and which is near the origin is

Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.

Area of the parallelogram whose sides are $x\cos \alpha + y\sin \alpha = p$ $x\cos \alpha + y\sin \alpha = q,\,\,$ $x\cos \beta + y\sin \beta = r$ and $x\cos \beta + y\sin \beta = s$ is

The co-ordinates of the orthocentre of the triangle bounded by the lines, $4x - 7y + 10 = 0; x + y=5$ and $7x + 4y = 15$ is :