The area of a triangle is $5$ and two of its vertices are $A(2, 1)$ and $B(3, -2)$. The third vertex,which lies on the line $y = x + 3$,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.

If the line $3x + 4y - 24 = 0$ intersects $X$ and $Y$ axes at points $A$ and $B$ respectively,then the incenter of the triangle $OAB$,where $O$ is the origin,is:

Two adjacent sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0$. If the equation of one diagonal is $11x + 7y - 9 = 0$,find the equation of the other diagonal.

The area of the triangle formed by the line $x \sin \alpha + y \cos \alpha = \sin 2\alpha$ and the coordinate axes is

The acute angle between the diagonals of a parallelogram whose vertices are $A(2, -1), B(0, 2), C(2, 3)$ and $D(4, 0)$ is