If $a, b$ are positive real numbers such that the lines $ax + 9y = 5$ and $4x + by = 3$ are parallel,then the least possible value of $a + b$ is

The line passing through the points $(1, 4)$ and $(-5, 1)$ intersects the line $4x + 3y - 5 = 0$ at the point:

The minimum area of the triangle formed by the variable line $3 \cos \theta \cdot x + 4 \sin \theta \cdot y = 12$ and the coordinate axes is

$A$ line $L$ perpendicular to the line $5x - 12y + 6 = 0$ makes a positive intercept on the $Y$-axis. If the distance from the origin to the line $L$ is $2$ units and the angle made by the perpendicular drawn from the origin to the line $L$ with the positive $X$-axis is $\theta$,then $\tan \theta + \cot \theta =$

Find the equation of the line whose perpendicular distance from the origin is $4$ units and the angle which the normal makes with the positive direction of the $x$-axis is $15^{\circ}$.

The equation of a line is $3x - 4y + 10 = 0$. Find its $x$-intercept and $y$-intercept.