If the point of intersection of the lines `2px+3qy+r=0` and `px-2qy-2r=0` lies strictly in the fourth quadrant and is equidistant from the two axes, then

Options

A. `5p+4q=0`
B. `4p-5q=0`
C. `4p+5q=0`
D. `5p-4q=0`

Answer

To solve the problem, we need to find the conditions under which the point of intersection of the lines \(2px + 3qy + r = 0\) and \(px - 2qy - 2r = 0\) lies strictly in the fourth quadrant and is equidistant from the two axes.

### Step 1: Find the point of intersection of the two lines.

To find the point of intersection, we can solve the two equations simultaneously.

1. The first equation is:
\[ 2px + 3qy + r = 0 \quad \text{(1)} \]
2. The second equation is:
\[ px - 2qy - 2r = 0 \quad \text{(2)} \]

We can express \(y\) from equation (1):
\[ 3qy = -2px - r \implies y = \frac{-2px - r}{3q} \quad \text{(3)} \]

Now substitute equation (3) into equation (2):
\[ px - 2q\left(\frac{-2px - r}{3q}\right) - 2r = 0 \]

### Step 2: Simplify the equation.

Substituting and simplifying:
\[ px + \frac{4px + 2r}{3} - 2r = 0 \]
Multiply through by 3 to eliminate the fraction:
\[ 3px + 4px + 2r - 6r = 0 \]
Combine like terms:
\[ 7px - 4r = 0 \implies 7px = 4r \implies x = \frac{4r}{7p} \quad \text{(4)} \]

### Step 3: Substitute \(x\) back to find \(y\).

Now substitute \(x\) from equation (4) back into equation (3):
\[ y = \frac{-2p\left(\frac{4r}{7p}\right) - r}{3q} = \frac{-\frac{8r}{7} - r}{3q} = \frac{-\frac{8r + 7r}{7}}{3q} = \frac{-\frac{15r}{7}}{3q} = \frac{-5r}{7q} \quad \text{(5)} \]

### Step 4: Analyze the coordinates.

Now we have the coordinates of the intersection point:
\[ \left( \frac{4r}{7p}, \frac{-5r}{7q} \right) \]

For the point to lie in the fourth quadrant:
- \(x > 0\) implies \(r > 0\) (since \(p\) is a constant and must not be zero).
- \(y < 0\) implies \(q > 0\) (since \(r\) is positive).

### Step 5: Condition of equidistance from the axes.

The point is equidistant from the axes if:
\[ \left| \frac{4r}{7p} \right| = \left| \frac{-5r}{7q} \right| \]
Since both \(r\) and \(p\) are positive, we can drop the absolute values:
\[ \frac{4r}{7p} = \frac{5r}{7q} \]
Cancelling \(r\) (since \(r \neq 0\)):
\[ \frac{4}{p} = \frac{5}{q} \implies 4q = 5p \implies 5p - 4q = 0 \]

### Conclusion

Thus, the final condition that satisfies the problem is:
\[ 5p - 4q = 0 \]