If $A$ and $B$ are two points on the line $3x + 4y + 15 = 0$ such that $OA = OB = 9$ units, then the area of the triangle $OAB$ is
$18 sq.$ units
$18\sqrt 2 sq.$units
18/$\sqrt 2 sq.$ units
None of these
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 :
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.
Area of the parallelogram formed by the lines ${a_1}x + {b_1}y + {c_1} = 0$,${a_1}x + {b_1}y + {d_1} = 0$and ${a_2}x + {b_2}y + {c_2} = 0$, ${a_2}x + {b_2}y + {d_2} = 0$is
If vertices of a parallelogram are respectively $(0, 0)$, $(1, 0)$, $(2, 2)$ and $(1, 2)$, then angle between diagonals is
A point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 13$. The equation of the locus of the point is