The line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in an anti-clockwise direction through an angle of $15^\circ$. The equation of the line in the new position is:

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$-axis,$y$-axis,and the $AB$ at $C, D,$ and $E$ respectively. If $O$ is the origin of coordinates,then the area of $OCEB$ is

Let $P$ be the point of intersection of the lines $L_1 \equiv x-y-7=0$ and $L_2 \equiv x+y-5=0$. $A(x_1, y_1)$ and $B(x_2, y_2)$ are points on the lines $L_1=0$ and $L_2=0$ respectively such that $PA=3\sqrt{2}$,$PB=\sqrt{2}$,$x_1, y_1 \geq 0$,$x_2, y_2 \geq 0$. Then the angle made by the line segment $AB$ at the origin is

$A$ line passing through $P(2,3)$ and making an angle of $30^{\circ}$ with the positive direction of the $x$-axis meets the curve $x^2-2xy-y^2=0$ at $A$ and $B$. Then the value of $PA \cdot PB$ is

If the equation of the base of an equilateral triangle is $2x - y = 1$ and the vertex is $(-1, 2)$,then the length of the side of the triangle is

$A$ straight line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$ square units. The equation of the line $L$ can be