$A$ square is formed by the lines $x=0, y=0, x=1, y=1$. Then,the equations of its diagonals will be

If $\Delta_1$ is the area of the triangle formed by the centroid and two vertices of a triangle,and $\Delta_2$ is the area of the triangle formed by the mid-points of the sides of the same triangle,then $\Delta_1 : \Delta_2 =$

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

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x + y = 14$ and $3x - 2y = 5$,respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is:

If the points $(a, b)$,$(a', b')$ and $(a - a', b - b')$ are collinear,then:

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