The angle between the lines joining the origin to the points of intersection of the line $x\sqrt{3} + y = 2$ and the curve $x^2 + y^2 = 4$ is

Consider the lines $L_1 \equiv 4x + 5y - 6 = 0$,$L_2 \equiv 2x + 3y - 4 = 0$,and $L_3 \equiv 3x - y + 2 = 0$. If the line $L_1 = 0$ intersects the lines $L_2 = 0$ and $L_3 = 0$ at the points $A$ and $B$ respectively,then the combined equation of lines $OA$ and $OB$ is

Two lines are given by $(x - 2y)^2 + k(x - 2y) = 0$. The value of $k$ so that the distance between them is $3$,is

The angle between the lines joining the points of intersection of the line $y = 3x + 2$ and the curve $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$ to the origin is:

$A$ pair of lines drawn through the origin forms a right-angled isosceles triangle with the line $2x + 3y = 6$,having the right angle at the origin. The area (in sq. units) of the triangle thus formed is

The condition that the lines joining the origin to the points of intersection of the line $\frac{x}{a} + \frac{y}{b} = 2$ and the circle $(x - a)^2 + (y - b)^2 = r^2$ are at right angles is