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

If $L$ is a line passing through the point $(-1, 1)$ and parallel to the common line of the pairs of lines $6x^2 - xy - 12y^2 = 0$ and $15x^2 + 14xy - 8y^2 = 0$,then the equation of the pair of lines joining the origin to the points of intersection of the curve $2x^2 - xy - y^2 + x - y = 0$ and the line $L$ is

Find the value$(s)$ of $k$ such that the distance between the two parallel lines represented by $(x-2y)^2 + k(x-2y) = 0$ is $3$ units.

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

$\triangle OAB$ is formed by the lines $x^2-4xy+y^2=0$ and the line $AB$. The equation of line $AB$ is $2x+3y-1=0$. Then the equation of the median of the triangle drawn from the origin is

The straight lines joining the origin to the points of intersection of the line $2x + y = 1$ and the curve $3x^2 + 4xy - 4x + 1 = 0$ include an angle of: