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

$A$ ray of light coming from the point $(1, 2)$ is reflected at a point $A$ on the $x$-axis and then passes through the point $(5, 3)$. The coordinates of the point $A$ are:

If $(h, k)$ is the image of the point $(3, -4)$ with respect to the line $2x - 3y - 5 = 0$ and $(\ell, m)$ is the foot of the perpendicular from $(h, k)$ onto the line $3x + 2y + 12 = 0$,then $\ell h + mk + 1 =$ ?

Let $Q$ be the image of a point $P(1, 2)$ with respect to the line $x + y + 1 = 0$ and $R$ be the image of $Q$ with respect to the line $x - y - 1 = 0$. If $M$ and $N$ are the midpoints of $PQ$ and $QR$ respectively,then $MN =$

The image of $A(1, -2)$ with respect to the straight line $L \equiv 2x - 3y + 5 = 0$ is $B$. The foot of the perpendicular from $P(-4, -1)$ onto the line joining $AB$ is:

The length of the perpendicular from the origin,on the normal to the curve $x^{2}+2xy-3y^{2}=0$ at the point $(2,2)$ is