The equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are
$x + \sqrt 3 y \pm 10 = 0$
$\sqrt 3 x + y \pm 10 = 0$
$x \pm \sqrt 3 y - 10 = 0$
None of these
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 :
If the extremities of the base of an isosceles triangle are the points $(2a,0)$ and $(0,a)$ and the equation of one of the sides is $x = 2a$, then the area of the triangle is
The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 \mathrm{x}+\mathrm{y}=14$ and $3 \mathrm{x}-2 \mathrm{y}=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. The equation of the side $\mathrm{BC}$ is :
The locus of the orthocentre of the triangle formed by the lines
$ (1+p) x-p y+p(1+p)=0, $
$ (1+q) x-q y+q(1+q)=0,$
and $y=0$, where $p \neq q$, is