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 :

The triangle $PQR$ is inscribed in the circle ${x^2} + {y^2} = 25$. If $Q$ and $R$ have co-ordinates $(3,4)$ and $(-4, 3)$ respectively, then $\angle QPR$ is equal to

Consider a triangle $\mathrm{ABC}$ having the vertices $\mathrm{A}(1,2), \mathrm{B}(\alpha, \beta)$ and $\mathrm{C}(\gamma, \delta)$ and angles $\angle \mathrm{ABC}=\frac{\pi}{6}$ and $\angle \mathrm{BAC}=\frac{2 \pi}{3}$. If the points $\mathrm{B}$ and $\mathrm{C}$ lie on the line $\mathrm{y}=\mathrm{x}+4$, then $\alpha^2+\gamma^2$ is equal to....................

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

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 :

The four points whose co-ordinates are $(2, 1), (1, 4), (4, 5), (5, 2)$ form :