Coordinates of the vertices of a quadrilateral are $(2, -1), (0, 2), (2, 3)$ and $(4, 0)$. The angle between its diagonals will be
${90^o}$
${0^o}$
${\tan ^{ - 1}}(2)$
${\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
If the line $3x + 3y -24 = 0$ intersects the $x-$ axis at the point $A$ and the $y-$ axis at the point $B$, then the incentre of the triangle $OAB$, where $O$ is the origin, is
The area of the triangle formed by the line $x\sin \alpha + y\cos \alpha = \sin 2\alpha $and the coordinates axes is
The area enclosed within the curve $|x| + |y| = 1$ is
The co-ordinates of the orthocentre of the triangle bounded by the lines, $4x - 7y + 10 = 0; x + y=5$ and $7x + 4y = 15$ is :
Without using the Pythagoras theorem, show that the points $(4,4),(3,5)$ and $(-1,-1)$ are vertices of a right angled triangle.