Why do we reverse inequality sign when dividing by negative number? We all learned in our early years that when dividing both sides by a negative number, we reverse the inequality sign.
Take
$-3x < 9$
To solve for $x$, we divide both sides by $-3$ and get
$$x > -3.$$
Why is the reversal of inequality? What is going in terms of number line that will help me understand the concept better?

Let $c$ be a negative number. In the case of multiplying both sides of an inequality by $c$, note that the function $f$ defined by $f(x) = cx$ is strictly decreasing on the entire real line. By definition, this means that if $x_{1} < x_{2},$ then $f\left(x_{1}\right) > f\left(x_{2}\right)$ (i.e. $cx_{1} > cx_{2}$). Incidentally, this is equivalent to $x_{2} > x_{1}$ implying $f\left(x_{2}\right) < f\left(x_{1}\right)$, so $f$ also reverses both types of strict inequalities. Moreover, it is not difficult to see that a strictly decreasing function reverses both types of non-strict inequalities. As for dividing both sides by a negative number, note that the function $g$ defined by $g(x) = \frac{1}{c}x$ is strictly decreasing on the entire real line. The same explanation can be used for taking the reciprocal of both sides of an inequality, when both sides are positive or when both sides are negative. In general, if a function $h$ is strictly decreasing on an interval $I$, then we can "take $h$" of both sides of an inequality as long as both sides belong to $I$ and we reverse the inequality. Similarly, strictly increasing functions preserve inequalities. This gives a sometimes useful application of the calculus task of determining on what interval(s) a function might be increasing or decreasing, by the way. For example, $\arctan(x)$ is strictly increasing on the entire real line, so you can take the arctangent of both sides of an inequality (keeping the inequality type the same).
