Let the triangle $PQR$ be the image of the triangle with vertices $(1,3), (3,1)$ and $(2,4)$ in the line $x+2y=2$. If the centroid of $\triangle PQR$ is the point $(\alpha, \beta)$,then $15(\alpha-\beta)$ is equal to:

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.

$A$ person standing at the junction (crossing) of $2$ straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$,wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. The equation of the path he should follow is:

If the image of point $P(2, 3)$ in a line $L$ is $Q(4, 5)$,then the image of point $R(0, 0)$ in the same line is

From the point $(-1, -1)$, two rays are sent making angles of $45^\circ$ with the line $x+y=0$. These rays get reflected from the mirror $x+2y=1$. If the equations of the reflected rays are $ax+by=9$ and $cx+dy=7$, where $a, b, c, d \in Z$, then the value of $ad+bc$ 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.......