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

Three points $P(h, k)$,$Q(x_{1}, y_{1})$ and $R(x_{2}, y_{2})$ lie on a line. Show that $(h-x_{1})(y_{2}-y_{1}) = (k-y_{1})(x_{2}-x_{1})$.

Find the equation of the line passing through the point $(2, 3)$ and perpendicular to the line joining the points $(-5, 6)$ and $(-6, 5)$.

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:

Find the equation of a straight line that makes an intercept of $3$ units on the negative direction of the $y$-axis and is inclined at an angle of ${\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$ with the $x$-axis.

If the normal form of the equation of a straight line $4x + 3y + 2 = 0$ is $x \cos \alpha + y \sin \alpha = p$ and its intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,then $\frac{p \sec \alpha}{ab} = $