$A$ 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 area of the triangle formed by the lines $x = 0$,$y = 0$,and $x/a + y/b = 1$ is:

Let $\alpha, \beta, \gamma, \delta \in \mathbb{Z}$ and let $A(\alpha, \beta), B(1, 0), C(\gamma, \delta)$ and $D(1, 2)$ be the vertices of a parallelogram $ABCD$. If $AB = \sqrt{10}$ and the points $A$ and $C$ lie on the line $3y = 2x + 1$,then $2(\alpha + \beta + \gamma + \delta)$ is equal to

If the lines $3x + y - 4 = 0$,$x - ay - 10 = 0$,and $bx + 2y + 9 = 0$ form three successive sides of a rectangle in that order and the fourth side passes through $(1, 2)$,then the area of that rectangle (in sq. units) is

Suppose a triangle is formed by $x+y=10$ and the coordinate axes. Then the number of points $(x, y)$ where $x$ and $y$ are natural numbers,lying inside the triangle is

If the lines $x + 3y = 4$ and $6x - 2y = 7$ are the diagonals of a parallelogram $PQRS$,then $PQRS$ is a: