The ends of the base of an isosceles triangle are at $(2a,\;0)$ and $(0,\;a).$ The equation of one side is $x=2a$ The equation of the other side is
$x + 2y - a = 0$
$x + 2y = 2a$
$3x + 4y - 4a = 0$
$3x - 4y + 4a = 0$
A line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( {\frac{1}{2},0} \right)$ and perpendicular to $L$. Then the area of the triangle formed by the lines $L,L'$ and $y$- axis, is
The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a
If the equation of the locus of a point equidistant from the points $({a_1},{b_1})$ and $({a_2},{b_2})$ is $({a_1} - {a_2})x + ({b_1} - {b_2})y + c = 0$, then the value of $‘c’$ is
Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$, then the equation of locus of mid- point $P$ of $BC$ is
Let the equations of two adjacent sides of a parallelogram $A B C D$ be $2 x-3 y=-23$ and $5 x+4 y$ $=23$. If the equation of its one diagonal $AC$ is $3 x +$ $7 y=23$ and the distance of A from the other diagonal is $d$, then $50 d ^2$ is equal to $........$.