$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point:

$A$ point moves in the $xy$-plane such that the sum of its distances from two mutually perpendicular lines is always equal to $5$ units. The area (in sq units) enclosed by the locus of the point is

Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

$A$ rod of length $8$ units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and $y+2=0$,respectively. If the locus of the point $P$,that divides the rod $AB$ internally in the ratio $2:1$ is $9(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=0$,then $\alpha-\beta-\gamma$ is equal to :

Suppose $P$ and $Q$ are the midpoints of the sides $AB$ and $BC$ of a triangle where $A(1, 3)$,$B(3, 7)$,and $C(7, 15)$ are vertices. Then the locus of $R$ satisfying $AC^2 + QR^2 = PR^2$ is