If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x$ $\cos \theta+ y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$ between the coordinates axes, then $\alpha$ is equal to
$7$
$-7$
$-7 \sqrt{3}$
$7 \sqrt{3}$
Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the equation to one diagonal is $11x + 7y = 9$, then the equation to the other diagonal 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 :
Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are