If vertices of a quadrilateral are $A(0,0), B(3,4), C(7,7)$ and $D(4,3)$,then the quadrilateral $ABCD$ is

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = \theta$,where $0 < \theta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $C$ is $7x - 4y - 1 = 0$,then $\tan \theta$ is equal to:

Let $PQR$ be a right-angled isosceles triangle,right-angled at $P(2, 1)$. If the equation of the side $QR$ is $2x + y = 3$,then the equation of one of the sides other than $QR$ is

The points $(0, 8/3), (1, 3)$ and $(82, 30)$ are:

Let $ABCD$ be a square of side length $1$. Let $P, Q, R, S$ be points in the interiors of the sides $AD, BC, AB, CD$ respectively,such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3\sqrt{3}}{4}$,then $RS$ equals

Without using the distance formula,show that the points $(-2,-1), (4,0), (3,3),$ and $(-3,2)$ are the vertices of a parallelogram.