We have two points:
- Point M: Located at coordinates $(p, q)$ in the first quadrant.
- Point N : Located at coordinates $(a, b)$ in the fourth quadrant.

The slope of the line joining $\(\mathbf{M}\)$ and $\(\mathbf{N}\)$ is positive. We need to compare:
- Quantity A: $a+b$
- Quantity B: $p+q$

And determine which is greater, or if they're equal, or if the relationship cannot be determined.

Properties of the Quadrants
First, let's recall the properties of the quadrants in the coordinate plane:
1. First Quadrant (I): Both $x$ and $y$ coordinates are positive.
- So, $p>0$ and $q>0$.
2. Fourth Quadrant (IV): $x$ coordinate is positive, $y$ coordinate is negative.
- So, $a>0$ and $b<0$.

Slope of the Line Joining M and N
The slope $m$ of the line joining two points ( $\(x_1, y_1\)$ ) and ( $\(x_2, y_2\)$ ) is given by:

$$
\(m=\frac{y_2-y_1}{x_2-x_1}\)
$$


Here, the slope is positive:

$$
\(\frac{b-q}{a-p}>0\)
$$


For a fraction to be positive, both the numerator and denominator must have the same sign (both positive or both negative).

Let's consider the possible cases:

Case 1: $b-q>0$ and $a-p>0$
- $b-q>0 \Rightarrow b>q$
- $a-p>0 \Rightarrow a>p$

But we know $b<0$ (since N is in IV) and $q>0$ (since M is in I), so $b>q$ would imply a negative number is greater than a positive number, which is impossible. Therefore, this case doesn't hold.

Case 2: $b-q<0$ and $a-p<0$
- $b-q<0 \Rightarrow b<q$
- $a-p<0 \Rightarrow a<p$

This is possible:
- $b<q$ : Since $b<0$ and $q>0$, this is always true.
- $a<p$ : Both $a$ and $p$ are positive, so this is possible.

So, the only valid scenario is:

$$
\(a<p \quad \text { and } \quad b<q\)
$$


Comparing Quantities A and B
Now, let's look at the quantities:
- Quantity A: $a+b$
- Quantity B: $p+q$

We know:
- $a<p$
- $b<q$ (and $b$ is negative, $q$ is positive)

Adding these two inequalities:

$$
\(a+b<p+q\)
$$


This directly implies that Quantity B is greater than Quantity A.