$A$ vertex of a square is $(3, 4)$ and one diagonal lies on the line $x + 2y = 1$. Find the equation of the second diagonal which passes through the given vertex.

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

Let $A(-2,-1)$,$B(1,0)$,$C(\alpha, \beta)$,and $D(\gamma, \delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x-y=5$ and the point $D$ lies on $3x-2y=6$,then the value of $|\alpha+\beta+\gamma+\delta|$ is equal to . . . . . . .

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

The point of intersection of the diagonals of the rectangle whose sides are contained in the lines $x=8, x=10, y=11$ and $y=12$ is

The area of the triangle formed by the lines $y = m_1x + c_1$,$y = m_2x + c_2$ and $x = 0$ is: