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

The equation of a straight line which passes through the point $(a \cos^3 \theta, a \sin^3 \theta)$ and is perpendicular to $x \sec \theta + y \operatorname{cosec} \theta = a$ is

Let $L$ be the line $y = 2x$ in the two-dimensional plane.
Statement $1$: The image of the point $(0, 1)$ in $L$ is the point $\left( \frac{4}{5}, \frac{3}{5} \right)$.
Statement $2$: The points $(0, 1)$ and $\left( \frac{4}{5}, \frac{3}{5} \right)$ lie on opposite sides of the line $L$ and are at equal distance from it.

$A$ light beam emanating from the point $A(3, 10)$ reflects from the line $2x + y - 6 = 0$ and then passes through the point $B(5, 6)$. The equations of the incident and reflected beams are respectively:

The image of the point $(4, -13)$ with respect to 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: