The perpendicular from the origin to a line meets it at the point $(-2, 9)$. Find the equation of the line.

The equation of the straight line passing through the point of intersection of the lines $5x - 6y - 1 = 0$ and $3x + 2y + 5 = 0$ and perpendicular to the line $3x - 5y + 11 = 0$ is

$A$ 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

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

The foot of the perpendicular from the point $(2,4)$ upon the line $x+y=4$ is:

The image of the point $P(3, 8)$ with respect to the line $x + 3y = 7$,assuming the line to be a plane mirror,is equal to .........