The angle between the line joining the points $(1, -2)$ and $(3, 2)$ and the line $x + 2y - 7 = 0$ is

The equation of a line making an angle $60^{\circ}$ with the line $x+y-3=0$ and passing through the point $(1,1)$ is

$A$ line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - p = 0, p > 0$,at the points $A$ and $B$,respectively. If $AB = \frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point $A$ on the line $L_2$ is $M$,then $\frac{AM}{BM}$ is equal to

$A$ straight line $L$ through the point $(3,-2)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis,then the equation of $L$ is

The angle between the lines $2x - y - 15 = 0$ and $3x + y + 4 = 0$ is .....$^o$.

If $\alpha, \beta$ $(\alpha > \beta)$ are two values of $k$ such that the equations $2x + (3 - 2k)y + (2k + 1) = 0$ and $kx + (k - 1)y - 4 = 0$ represent two perpendicular lines,then $\alpha^2 + 2\beta =$