Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line $2x + y = 5$ . Then the area of the triangle is :

Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4,5),(0,7) (5,-5)$ and $(-4,-2) .$ Also, find its area.

A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is

The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 \mathrm{x}+\mathrm{y}=14$ and $3 \mathrm{x}-2 \mathrm{y}=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. The equation of the side $\mathrm{BC}$ is :

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

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.