The equation of the line whose slope is $3$ and which cuts off an intercept of $3$ from the positive $x$-axis is

$A$ man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is $\frac{1}{4}$. Three stones $A, B$ and $C$ are placed at the points $(1,1), (2,2)$ and $(4,4)$ respectively. Which of these stones is/are on the path of the man?

For what values of $a$ and $b$ are the intercepts cut off on the coordinate axes by the line $ax + by + 8 = 0$ equal in length but opposite in sign to those cut off by the line $2x - 3y + 6 = 0$ on the axes?

$A$ straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y$-axis respectively. If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with the positive direction of $y$-axis and the area of $\triangle OAB$ is $\frac{98}{3} \sqrt{3}$,then $a^2 - b^2$ is equal to:

Reduce the following equation into slope-intercept form and find its slope and the $y$-intercept: $6x + 3y - 5 = 0$.

The equation of the line passing through the origin and the point $(a \cos \theta, a \sin \theta)$ is .....