The image of the point $(6, 5)$ in the line $x = 3$ is

$A$ ray of light passing through the point $P(2, 3)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $Q(5, 4)$. Let $R$ be the point that divides the line segment $AQ$ internally in the ratio $2:1$. Let the coordinates of the foot of the perpendicular $M$ from $R$ on the bisector of the angle $PAQ$ be $(\alpha, \beta)$. Then,the value of $7\alpha + 3\beta$ is equal to.......

If the foot of the perpendicular drawn from the origin to a straight line is $(3, -4)$,then the equation of the line is:

$A$ ray of light is incident along a line which meets another line,$7x - y + 1 = 0$,at the point $(0, 1)$. The ray is then reflected from this point along the line,$y + 2x = 1$. Then the equation of the line of incidence of the ray of light is

Let $a$ and $b$ be any two numbers satisfying $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$. Then,the foot of the perpendicular from the origin on the variable line $\frac{x}{a} + \frac{y}{b} = 1$ lies on

The perpendicular bisector of the line segment joining $P(1, 4)$ and $Q(k, 3)$ has a $y-$intercept of $-4$. Then a possible value of $k$ is: