$A$ vertex of an equilateral triangle is $(2, 3)$ and the equation of the opposite side is $x + y = 2$. Then,the equation of one of the other two sides is:

If the sides $AB, BC, CD$ and $DA$ of a quadrilateral are given by the equations $x + 2y = 3, x = 1, x - 3y = 4$ and $5x + y + 12 = 0$ respectively,then find the angle between the diagonals $AC$ and $BD$ in degrees.

If the incentre of the triangle formed by the lines $x-2=0$,$x+y-1=0$,and $x-y+3=0$ is $(\alpha, \beta)$,then $\beta=$

One side of a square is inclined at an acute angle $\alpha$ with the positive $x-$axis,and one of its extremities is at the origin. If the remaining three vertices of the square lie above the $x-$axis and the side of a square is $4$,then the equation of the diagonal of the square which is not passing through the origin is

The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of:

The area (in square units) of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is: