The point $(2,3)$ is first reflected in the straight line $y=x$ and then translated through a distance of $2$ units along the positive direction of the $X$-axis. The coordinates of the transformed point are

The value of $a$ for which the image of the point $(a, a - 1)$ with respect to the line mirror $3x + y = 6a$ is the point $(a^2 + 1, a)$ is:

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 $A \left(\frac{3}{\sqrt{a}}, \sqrt{a}\right)$ 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:

Let $L$ denote the line in the $xy$-plane with $x$ and $y$ intercepts as $3$ and $1$ respectively. Then the image of the point $(-1, -4)$ in this line is

If the image of point $P(\lambda, 2\lambda)$ in the line mirror $x - y + 2 = 0$ is $Q(\mu, 3\lambda)$,then choose the correct option.