The coordinates of the vertices of a quadrilateral are $(2, -1)$,$(0, 2)$,$(2, 3)$,and $(4, 0)$. The angle between its diagonals is:

The area (in sq units) of the quadrilateral formed by the lines $2x + 3y + 6 = 0$,$2x - 3y + 6 = 0$,$2x + 3y - 6 = 0$,and $2x - 3y - 6 = 0$ is

The points $\left( \frac{a}{\sqrt{3}}, a \right)$,$\left( \frac{2a}{\sqrt{3}}, 2a \right)$,and $\left( \frac{a}{\sqrt{3}}, 3a \right)$ are the vertices of:

Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15x - y = 82$,$6x - 5y = -4$,and $9x + 4y = 17$. Then $\alpha + 2\beta$ and $2\alpha - \beta$ are the roots of the equation $...........$.

In the triangle with vertices at $A(6,3), B(-6,3)$ and $C(-6,-3)$,the median through $A$ meets $BC$ at $P$,the line $AC$ meets the $x$-axis at $Q$,while $R$ and $S$ respectively denote the orthocentre and centroid of the triangle. Then the correct matching of the coordinates of points in List-$I$ to List-$II$ is:

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

Suppose a triangle is formed by $x+y=10$ and the coordinate axes. Then the number of points $(x, y)$ where $x$ and $y$ are natural numbers,lying inside the triangle is