The line `(p+2q)x+(p-3q)y=p-q` for different values of p and q passes through the point

Options

A. `(3//2,5//2)`
B. `(2//5,2//5)`
C. `(3//5,3//5)`
D. `(2//5,3//5)`

Answer

To solve the problem, we need to determine the values of \(x\) and \(y\) for the line given by the equation:

\[ (p + 2q)x + (p - 3q)y = p - q \]

This line must pass through a specific point for different values of \(p\) and \(q\).

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate terms involving \(p\) and \(q\):

\[ (p + 2q)x + (p - 3q)y - (p - q) = 0 \]

### Step 2: Grouping Terms
We can group the terms involving \(p\) and \(q\):

\[ px + py + 2qx - 3qy - p + q = 0 \]

This can be rewritten as:

\[ p(x + y - 1) + q(2x - 3y + 1) = 0 \]

### Step 3: Setting Coefficients to Zero
For the equation to hold for all values of \(p\) and \(q\), the coefficients of \(p\) and \(q\) must both equal zero:

1. \(x + y - 1 = 0\) (Equation 1)
2. \(2x - 3y + 1 = 0\) (Equation 2)

### Step 4: Solving the System of Equations
Now we solve the system of equations formed by Equation 1 and Equation 2.

From Equation 1:

\[ x + y = 1 \quad \text{(1)} \]

From Equation 2:

\[ 2x - 3y = -1 \quad \text{(2)} \]

### Step 5: Substituting for \(y\)
We can express \(y\) in terms of \(x\) using Equation 1:

\[ y = 1 - x \]

### Step 6: Substituting into Equation 2
Now substitute \(y\) into Equation 2:

\[ 2x - 3(1 - x) = -1 \]

Expanding this gives:

\[ 2x - 3 + 3x = -1 \]

Combining like terms:

\[ 5x - 3 = -1 \]

### Step 7: Solving for \(x\)
Now, solve for \(x\):

\[ 5x = 2 \implies x = \frac{2}{5} \]

### Step 8: Finding \(y\)
Substituting \(x = \frac{2}{5}\) back into Equation 1 to find \(y\):

\[ \frac{2}{5} + y = 1 \implies y = 1 - \frac{2}{5} = \frac{3}{5} \]

### Final Answer
Thus, the point through which the line passes is:

\[ \left(\frac{2}{5}, \frac{3}{5}\right) \]