1. The line $\(\ell\)$ passes through the point $(a, b)$.
2. The slope of line $\(\ell\)$ is $\(\frac{3}{2}\)$.
3. The line $\(\ell\)$ passes through the origin $(0,0)$.

Since the line passes through the origin $(0,0)$ and the point $(a, b)$, we can use the slope formula:
Slope $\(m=\frac{y_2-y_1}{x_2-x_1}\)$
Using $\(\left(x_1, y_1\right)=(0,0)\)$ and $\(\left(x_2, y_2\right)=(a, b)\)$ :

$$
\(m=\frac{b-0}{a-0}=\frac{b}{a}\)
$$


We are given that the slope $\(m=\frac{3}{2}\)$.
So, $\(\frac{b}{a}=\frac{3}{2}\)$.
This equation establishes a relationship between $a$ and $b$.
From $\(\frac{b}{a}=\frac{3}{2}\)$, we can cross-multiply to get $2 b=3 a$.
This can also be written as $b=\frac{3}{2} a$.
Now, let's compare Quantity A (a) and Quantity B (b).
We have $\(b=\frac{3}{2} a\)$.
Consider the possible cases for $a$ :

Case 1: $a$ is a positive number.
If $a>0$, then $\(b=\frac{3}{2} a\)$ will also be a positive number.

Since $\(\frac{3}{2}=1.\)5$, we have $b=1.5 a$.

In this case, $b$ is 1.5 times $a$.

For example, if $a=2$, then $\(b=1.5 \times 2=3\)$. Here, $b>a$.

If $a=10$, then $\(b=1.5 \times 10=15\)$. Here, $b>a$.

Case 2: $a$ is a negative number.

If $a<0$, then $\(b=\frac{3}{2} a\)$ will also be a negative number.

Since $\(\frac{3}{2}=1.5\)$, we have $b=1.5 a$.

In this case, $b$ is 1.5 times $a$.

For example, if $a=-2$, then $\(b=1.5 \times(-2)=-3\)$. Here, $b<a$ (since -3 is less than -2 ).

If $a=-10$, then $\(b=1.5 \times(-10)=-15\)$. Here, $b<a$ (since -15 is less than -10 ).

Since the relationship between $a$ and $b$ depends on whether $a$ is positive or negative, we cannot definitively say whether $a$ is greater than $b$ or $b$ is greater than $a$.

- If $a>0$, then $b>a$.
- If $a<0$, then $b<a$.

Therefore, the relationship between Quantity A and Quantity B cannot be determined.