The equation of the locus of a point which is equidistant from the points $(2, 3)$ and $(4, 5)$ is

The equation of the locus of the point of intersection of the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ is

Find the fixed point through which the line $x(a + 2b) + y(a + 3b) = a + b$ always passes for all values of $a$ and $b$.

$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

Let $P(2, 3)$,$Q(6, 0)$,and $R(\alpha, \beta)$ be three points in the $x-y$ plane such that $|PR + QR| + |PR - QR|$ is minimum. Then the value of $(\alpha - 2\beta)$ is:

$B(2,3)$,$C(5,-2)$,and $D(1,-1)$ are three points. If $A(x, y)$ is a variable point such that the area of the quadrilateral $ABCD$ is $10 \text{ sq. units}$,then the locus of $A$ is