Find the equation of the straight line passing through $(0,0)$ and the foot of the perpendicular from $(2,4)$ onto the line $x+y-1=0$.

Sum of values of $k$ for which the image of the point $(\lambda^2 + 1, \lambda)$ with respect to the line $y = -3x + 6k$ is the point $(\lambda, \lambda - 1)$ is:

The reflection of the point $(4, -13)$ in the line $5x + y + 6 = 0$ is

$A(-1, 1)$ and $B(5, 3)$ are opposite vertices of a square in the $xy$-plane. The equation of the other diagonal (not passing through $A$ and $B$) of the square is given by:

Let a ray of light passing through the point $(3,10)$ reflect on the line $2x+y=6$ and the reflected ray pass through the point $(7,2)$. If the equation of the incident ray is $ax+by+1=0$,then $a^2+b^2+3ab$ is equal to:

$A$ light ray emits from the origin making an angle $30^{\circ}$ with the positive $x$-axis. After getting reflected by the line $x + y = 1$,if this ray intersects the $x$-axis at $Q$,then the abscissa of $Q$ is