The equation of the line perpendicular to $2x - 3y + 5 = 0$ and making an intercept of $3$ with the positive $Y$-axis is

The equations of the sides $AB$,$AC$,and $BC$ of a $\triangle ABC$ are respectively $x-3y=0$,$3x-y=0$,and $x+y+4=0$. If $P$ and $Q$ are points on the line $3x-y+k=0$ passing through $B$ such that $PB:BQ=1:1$,then $k=$

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points $(2, -3)$ and $(4, -5)$,then $(a, b) = $

If the line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in anticlockwise direction through an angle of $15^{\circ}$,then the equation of the line in the new position is

The polar equation of the line perpendicular to the line $\sin \theta - \cos \theta = \frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6}\right)$ is

The line joining two points $A(2, 0)$ and $B(3, 1)$ is rotated about $A$ in the anticlockwise direction through an angle of $15^{\circ}$. The equation of the line in the new position is: