Find the $y$-intercept of a line that is perpendicular to $3x + y = 3$ and passes through the point $(2, 2)$.

Find the value of $k$ for which the line $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$ is parallel to the $y$-axis.

If the three points are $A(2, 3)$,$B(3, 1)$,and $C(5, 3)$,then the slope of the line passing through $A$ and bisecting $BC$ is:

The intercept of a line between the coordinate axes is divided by the point $(-5, 4)$ in the ratio $1 : 2$. The equation of the line will be:

In what direction should a line be drawn through the point $(1, 2)$ so that its point of intersection with the line $x + y = 4$ is at a distance of $\frac{\sqrt{6}}{3}$ from the given point (in $^\circ$)?

If $x \cos \theta + y \sin \theta = p$ is the normal form and $y = mx + c$ is the slope-intercept form of the line $x + 2y + 1 = 0$,then $\tan^{-1}(\tan \theta + m + c) = $