$A$ 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

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

The equation of the line perpendicular to the line $ax + by + c = 0$ and passing through $(a, b)$ is:

Let $A \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$ with $a > 0$ be a fixed point in the $xy$-plane. The image of $A$ in the $y$-axis is $B$,and the image of $B$ in the $x$-axis is $C$. If $D(3 \cos \theta, a \sin \theta)$ is a point in the fourth quadrant such that the maximum area of $\triangle ACD$ is $12$ square units,then $a$ is equal to:

Find the coordinates of the foot of the perpendicular drawn from the point $(7, 8)$ to the line $2x + 3y - 4 = 0$.

The distance from the origin to the image of $(1,1)$ with respect to the line $x+y+5=0$ is (in $\sqrt{2}$)