The point of injtersection of the line `x/p+y/q=1 and x/q+y/p=1` lies on the line

Options

A. `x-y=0`
B. `(x+y)(p+q)=2qp`
C. `(ax+by)(p+q)=(a+b)pq`
D. `(ax-by)(p+q)=)a-b)pq`

Answer

To solve the problem of finding the point of intersection of the lines given by the equations \( \frac{x}{p} + \frac{y}{q} = 1 \) and \( \frac{x}{q} + \frac{y}{p} = 1 \), we can follow these steps:

### Step 1: Write down the equations
We start with the two equations:
1. \( \frac{x}{p} + \frac{y}{q} = 1 \) (Equation 1)
2. \( \frac{x}{q} + \frac{y}{p} = 1 \) (Equation 2)

### Step 2: Eliminate one variable
To find the point of intersection, we can subtract Equation 2 from Equation 1. This will help us eliminate one of the variables.

Subtracting Equation 2 from Equation 1:
\[ \left( \frac{x}{p} - \frac{x}{q} \right) + \left( \frac{y}{q} - \frac{y}{p} \right) = 0 \]

### Step 3: Simplify the equation
Now, we can simplify the left-hand side:
\[ \frac{x}{p} - \frac{x}{q} = x \left( \frac{1}{p} - \frac{1}{q} \right) = x \left( \frac{q - p}{pq} \right) \]
\[ \frac{y}{q} - \frac{y}{p} = y \left( \frac{1}{q} - \frac{1}{p} \right) = y \left( \frac{p - q}{pq} \right) \]

So we have:
\[ x \left( \frac{q - p}{pq} \right) + y \left( \frac{p - q}{pq} \right) = 0 \]

### Step 4: Factor out common terms
Factoring out \( \frac{1}{pq} \) (which is non-zero), we get:
\[ (q - p)x + (p - q)y = 0 \]

### Step 5: Rearranging the equation
Rearranging gives us:
\[ (q - p)x = (q - p)y \]

If \( q \neq p \), we can divide both sides by \( (q - p) \):
\[ x = y \]

### Conclusion
Thus, the point of intersection of the two lines lies on the line \( x = y \).