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

A pair of straight lines drawn through the origin form with the line $2x + 3y = 6$ an isosceles right angled triangle, then the lines and the area of the triangle thus formed is

In an isosceles triangle $ABC$, the coordinates of the points $B$ and $C$ on the base $BC$ are respectively $(1, 2)$ and $(2, 1)$. If the equation of the line $AB$ is $y = 2x$, then the equation of the line $AC$ is

The sides $AB,BC,CD$ and $DA$ of a quadrilateral are $x + 2y = 3,\,x = 1,$ $x - 3y = 4,\,$ $\,5x + y + 12 = 0$ respectively. The angle between diagonals $AC$ and $BD$ is ......$^o$

Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4,5),(0,7) (5,-5)$ and $(-4,-2) .$ Also, find its area.

If the coordinates of the vertices of the triangle $ABC$ be $(-1, 6)$, $(-3, -9)$, and $(5, -8)$ respectively, then the equation of the median through $C$ is