Suppose a point $P$ moves so that $BP^2 - AP^2 = 121$,where $A$ and $B$ are $(2, 5)$ and $(5, 11)$ respectively. Then the locus of $P$ is a straight line,whose slope is

If the sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$,show that $P$ must move on a line.

Let $A=(2,3)$ and $B=(3,-5)$ be two vertices of $\triangle ABC$ such that $C$ is a point on the line $L \equiv 3x + 4y - 5 = 0$. Then the locus of the centroid of $\triangle ABC$ is a line parallel to

Let $A(-1,1)$ and $B(2,3)$ be two points and $P(x,y)$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$,then $5a+2b$ is:

$A$ man starts walking from the point $P(-3, 4)$,touches the $x$-axis at $R$,and then turns to reach the point $Q(0, 2)$. The man is walking at a constant speed. If the man reaches the point $Q$ in the minimum time,then $50((PR)^{2} + (RQ)^{2})$ is equal to ..... .

If the sum of the distances of a point $P(x, y)$ from two perpendicular lines in a plane is $1$,then the locus of $P$ is a