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 :

The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is 

If the straight line $ax + by + c = 0$ always passes through $(1, -2),$ then $a, b, c$ are

The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is :

An equilateral triangle has each of its sides of length $6\,\, cm$ . If $(x_1, y_1) ; (x_2, y_2) \,\, and \,\, (x_3, y_3)$ are its vertices then the value of the determinant,${{\left| {\,\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}\,} \right|}^2}$ is equal to :

Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :