The equation of the line passing through $(1, 1)$ and parallel to the line $2x + 3y - 7 = 0$ is

Let $PS$ be the median of the triangle with vertices $P(2, 2)$,$Q(6, -1)$,and $R(7, 3)$. The equation of the line passing through $(1, -1)$ and parallel to $PS$ is

The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$,then $\frac{b}{a}$ is equal to

Find the equation of a line drawn perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ through the point where it meets the $y-$ axis.

The equation of the line passing through the points $(1, 2)$ and $(2, 5)$ is:

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is