The coordinates of the foot of the perpendicular drawn from the point $(-2, 3)$ to the line $3x - y - 1 = 0$ are:

The orthocentre and the centroid of $\triangle ABC$ are $(5,8)$ and $\left(3, \frac{14}{3}\right)$ respectively. The equation of the side $BC$ is $x-y=0$. Given that the image of the orthocentre of a triangle with respect to any side lies on the circumcircle of that triangle,then the diameter of the circumcircle of $\triangle ABC$ is

Let $\alpha$ and $\beta$ be integers satisfying $0 < \beta < \alpha$. Let $P(\alpha, \beta)$ be a point. Let $Q$ be the reflection of $P$ in the line $y = x$,$R$ be the reflection of $Q$ in the $y$-axis,$S$ be the reflection of $R$ in the $x$-axis,and $T$ be the reflection of $S$ in the $y$-axis. If the area of the convex pentagon $PQRST$ is $187 \ sq. \ units$,then the value of $\alpha + \beta^2$ is:

The point $Q$ is the image of the point $P(1,5)$ about the line $y=x$ and $R$ is the image of the point $Q$ about the line $y=-x$. The circumcentre of the $\Delta PQR$ is

The equation of the 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

$A(-1, 1)$ 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: