The reflection of the point $(1, 1)$ along the line $y = -x$ is

The point $(a, b)$ is the foot of the perpendicular drawn from the point $(3, 1)$ to the line $x + 3y + 4 = 0$. If $(p, q)$ is the image of $(a, b)$ with respect to the line $3x - 4y + 11 = 0$,then $\frac{p}{a} + \frac{q}{b} =$

$A$ ray of light along $x + \sqrt{3}y = \sqrt{3}$ gets reflected upon reaching the $x$-axis. The equation of the reflected ray is:

Let $P$ be an interior point of a $\triangle ABC$. Let $Q$ and $R$ be the reflections of $P$ in $AB$ and $AC$,respectively. If $Q, A, R$ are collinear,then $\angle A$ equals (in $^{\circ}$)

Let $P$ be a point inside a $\triangle ABC$ with $\angle ABC = 90^{\circ}$. Let $P_1$ and $P_2$ be the images of $P$ under reflection in $AB$ and $BC$ respectively. The distance between the circumcenters of $\triangle ABC$ and $\triangle P_1PP_2$ is

The equation of a line passing through $(1, -2)$ and perpendicular to the line $3x - 5y + 7 = 0$ is