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

The equation of the perpendicular bisector of the line segment joining $A(-2, 3)$ and $B(6, -5)$ is

Let $0 < \alpha < \pi/2$ be a constant angle. If $P \equiv (\cos \theta, \sin \theta)$ and $Q \equiv (\cos(\alpha - \theta), \sin(\alpha - \theta))$,how is $Q$ obtained from $P$?

Sum of values of $k$ for which the image of the point $(\lambda^2 + 1, \lambda)$ with respect to the line $y = -3x + 6k$ is the point $(\lambda, \lambda - 1)$ 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 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 .........