The co-ordinates of three points $A(-4, 0) ; B(2, 1)$ and $C(3, 1)$ determine the vertices of an equilateral trapezium $ABCD$ . The co-ordinates of the vertex $D$ are :
$(6, 0)$
$(- 3, 0)$
$(- 5, 0)$
$(9, 0)$
$ABC$ is an isosceles triangle . If the co-ordinates of the base are $(1, 3)$ and $(- 2, 7) $, then co-ordinates of vertex $A$ can be :
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
Let $b, d>0$. The locus of all points $P(r, \theta)$ for which the line $P$ (where, $O$ is the origin) cuts the line $r \sin \theta=b$ in $Q$ such that $P Q=d$ is
If the equation of base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$, then the length of the side of the triangle is
If one vertex of an equilateral triangle of side $'a'$ lies at the origin and the other lies on the line $x - \sqrt{3} y = 0$ then the co-ordinates of the third vertex are :