Equation of one of the sides of an isosceles right angled triangle whose hypotenuse is $3x + 4y = 4$ and the opposite vertex of the hypotenuse is $(2, 2)$, will be
$x - 7y + 12 = 0$
$7x + y - 12 = 0$
$x - 7y + 16 = 0$
$y - (3 - 2\sqrt 2 )x = 0$
The equation of the line which makes right angled triangle with axes whose area is $6$ sq. units and whose hypotenuse is of $5$ units, 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.
For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distance of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P)+d_2(P) \leq 4$, is
The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is
Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are