The two vertices of a triangle are $(2, -1)$ and $(3, 2)$,and the third vertex lies on the line $x + y = 5$. If the area of the triangle is $4$ square units,then the third vertex is:

$A$ square with each side of length $a$ lies above the $x$-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle $\alpha$ $(0 < \alpha < \frac{\pi}{4})$ with the positive direction of the $x$-axis. Find the equations of the diagonals of the square.

$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is

The coordinates of three points $A(-4, 0)$,$B(2, 1)$,and $C(3, 1)$ determine the vertices of an isosceles trapezium $ABCD$. The coordinates of the vertex $D$ are:

Find the area of the quadrilateral formed by the lines $4y - 3x = 1, 4y - 3x - 3 = 0, 3y - 4x + 1 = 0,$ and $3y - 4x + 2 = 0$.

The diagonal passing through the origin of a quadrilateral formed by $x = 0, y = 0, x + y = 1$ and $6x + y = 3$ is