Consider a triangle $ABC$ having the vertices $A(1,2)$,$B(\alpha, \beta)$,and $C(\gamma, \delta)$. The angles are $\angle ABC = \frac{\pi}{6}$ and $\angle BAC = \frac{2\pi}{3}$. If the points $B$ and $C$ lie on the line $y = x + 4$,then $\alpha^2 + \gamma^2$ is equal to:

The slope of a line is double the slope of another line. If the tangent of the angle between them is $\frac{1}{3},$ find the slopes of the lines.

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

Let $\theta_1$ be the angle between two lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_2 = 0$,and $\theta_2$ be the angle between two lines $2x + 3y + c_1 = 0$ and $-x + 5y + c_3 = 0$,where $c_1, c_2, c_3$ are any real numbers.
Statement-$1$: If $c_2$ and $c_3$ are proportional,then $\theta_1 = \theta_2$.
Statement-$2$: $\theta_1 = \theta_2$ for all $c_2$ and $c_3$.

The lines $y = 2x$ and $x = -2y$ are