The origin and the points where the line $L_1$ intersects the $x$-axis and $y$-axis are vertices of a right-angled triangle $T$ whose area is $8$. Also,the line $L_1$ is perpendicular to the line $L_2: 4x - y = 3$. Then,the perimeter of triangle $T$ is:

$A$ is a point on either of two lines $y + \sqrt{3} |x| = 2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of the perpendicular from $A$ on the bisector of the angle between them are

$A$ straight line passing through the origin $O$ intersects the lines $10x - 8y - 10 = 0$ and $\frac{x}{4} - \frac{y}{5} + 1 = 0$ at right angles at points $P$ and $Q$ respectively. Then the ratio in which $O$ divides the line segment $PQ$ is

The line $x + y = p$ meets the $x$ and $y$ axes at $A$ and $B$ respectively. $A$ triangle $APQ$ is inscribed in the triangle $OAB$,where $O$ is the origin,with a right angle at $Q$. $P$ and $Q$ lie on $OB$ and $AB$ respectively. If the area of triangle $APQ$ is $3/8$ of the area of triangle $OAB$,then $\frac{AQ}{BQ}$ is equal to:

$A$ square of side $a$ lies above the $x$-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha, (0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. The equation of its diagonal not passing through the origin is

The number of integral points (integral point means both the coordinates should be integers) exactly in the interior of the triangle with vertices $(0, 0)$,$(0, 21)$,and $(21, 0)$ is: