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.

Let $L$ be the line joining the origin to the point of intersection of the lines represented by $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. If $L$ is perpendicular to the line $kx + y + 3 = 0$,then $k$ is equal to

The equation of the straight line joining the origin to the point of intersection of $y - x + 7 = 0$ and $y + 2x - 2 = 0$ 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

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

The perpendicular distance between the lines given by $(x-2y+1)^2 + k(x-2y+1) = 0$ is $\sqrt{5}$,then $k=$