If $Q$ is the image of the point $P(1,1)$ with respect to the straight line $x+y+1=0$,then the length of the perpendicular drawn from $Q$ to the line $3x-4y+3=0$ is

Let $\alpha$ and $\beta$ be integers satisfying $0 < \beta < \alpha$. Let $P(\alpha, \beta)$ be a point. Let $Q$ be the reflection of $P$ in the line $y = x$,$R$ be the reflection of $Q$ in the $y$-axis,$S$ be the reflection of $R$ in the $x$-axis,and $T$ be the reflection of $S$ in the $y$-axis. If the area of the convex pentagon $PQRST$ is $187 \ sq. \ units$,then the value of $\alpha + \beta^2$ is:

The image of a point $A(3, 8)$ in the line $x + 3y - 7 = 0$ is

The equation of the perpendicular bisector of the line segment joining the points $(7, 4)$ and $(-1, -2)$ is:

The point $(a, b)$ is the foot of the perpendicular drawn from the point $(3, 1)$ to the line $x + 3y + 4 = 0$. If $(p, q)$ is the image of $(a, b)$ with respect to the line $3x - 4y + 11 = 0$,then $\frac{p}{a} + \frac{q}{b} =$

The equation of the line perpendicular to the line $ax + by + c = 0$ and passing through $(a, b)$ is: