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

$A$ straight line passing through a fixed point $(3, 5)$ intersects the coordinate axes at two points $A$ and $B$. If the locus of $C(x, y)$,which forms a rectangle with the points $A, O$ (origin),and $B$,is $ax + 2hxy + by = 0$,then $a + b + h =$

$A$ line $L$ passing through the point $P(-5, -4)$ cuts the lines $x-y-5=0$ and $x+3y+2=0$ at $Q$ and $R$ respectively such that $\frac{18}{PQ} + \frac{15}{PR} = 2$. Then the slope of the line $L$ is:

Given $\frac{x}{a} + \frac{y}{b} = 1$ and $ax + by = 1$ are two variable lines,where $a$ and $b$ are parameters connected by the relation $a^2 + b^2 = ab$. The locus of the point of intersection has the equation:

If the line $\frac{x}{a} + \frac{y}{b} = 1$ is a variable line such that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,then the locus of the foot of the perpendicular from the origin to the line is:

If the point $(x, y)$ is equidistant from the points $(a + b, b - a)$ and $(a - b, a + b)$,then: