The equation of the line passing through $(1, 2)$ and perpendicular to $3x + 4y + 5 = 0$ is

The equation of the line,which bisects the line segment joining the points $(2, -19)$ and $(6, 1)$ and is perpendicular to the line joining the points $(-1, 3)$ and $(5, -1)$,is

Let $\alpha \in R$. If the line $(\alpha+1) x+\alpha y+\alpha=1$ passes through a fixed point $(h, k)$ for all $\alpha$,then $h^2+k^2=$

Find the slope of the line making an inclination of $60^{\circ}$ with the positive direction of the $x$-axis.

If we reduce $3x + 3y + 7 = 0$ to the form $x \cos \alpha + y \sin \alpha = p$,then the value of $p$ is

$A$ straight line makes an angle of $135^\circ$ with the $x$-axis and cuts the $y$-axis at a distance of $-5$ from the origin. The equation of the line is