The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points $(a^2 + 1 , a^2 + 1 )$ and $(2a, - 2a)$, $a \ne 0$. Then for any $a$ , the orthocentre of this triangle lies on the line

Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.

$A(-1, 1)$, $B(5, 3)$ are opposite vertices of a square in $xy$-plane. The equation of the other diagonal (not passing through $(A, B)$ of the square is given by

If $x^2-y^2+2 h x y+2 g x+2 f y+c=0$ is the locus of a point, which moves such that it is always equidistant from the lines $x+2 y+7=0$ and $2 x-y$ $+8=0$, then the value of $\mathrm{g}+\mathrm{c}+\mathrm{h}-\mathrm{f}$ equals

Let the circumcentre of a triangle with vertices $A ( a , 3), B ( b , 5)$ and $C ( a , b ), ab >0$ be $P (1,1)$. If the line $AP$ intersects the line $BC$ at the point $Q \left( k _{1}, k _{2}\right)$, then $k _{1}+ k _{2}$ is equal to.

The area of triangle formed by the lines $x + y - 3 = 0 , x - 3y + 9 = 0$ and $3x - 2y + 1= 0$