the lines `(p+2q)x+(p-3q)y=p-q` for different values of `p&q` passes trough the fixed point is:

Options

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

Answer

To find the fixed point through which the lines given by the equation \((p + 2q)x + (p - 3q)y = p - q\) pass for different values of \(p\) and \(q\), we can follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ (p + 2q)x + (p - 3q)y = p - q \]
We can rearrange this equation to bring all terms to one side:
\[ (p + 2q)x + (p - 3q)y - (p - q) = 0 \]

### Step 2: Expand and group terms
Expanding the equation, we have:
\[ px + 2qx + py - 3qy - p + q = 0 \]
Now, we can group the terms involving \(p\) and \(q\):
\[ px + py + 2qx - 3qy - p + q = 0 \]

### Step 3: Factor out common terms
We can factor the equation as follows:
\[ p(x + y - 1) + q(2x - 3y + 1) = 0 \]

### Step 4: Set the coefficients to zero
For the equation to hold for all values of \(p\) and \(q\), both coefficients must equal zero:
1. \(x + y - 1 = 0\) (Equation 1)
2. \(2x - 3y + 1 = 0\) (Equation 2)

### Step 5: Solve the system of equations
Now we will solve the two equations simultaneously.

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

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

### Step 6: Substitute \(y\) from Equation 1 into Equation 2
From Equation (1), we can express \(y\) in terms of \(x\):
\[ y = 1 - x \]
Substituting this into Equation (2):
\[ 2x - 3(1 - x) = -1 \]
Expanding this gives:
\[ 2x - 3 + 3x = -1 \]
Combining like terms:
\[ 5x - 3 = -1 \]
Adding 3 to both sides:
\[ 5x = 2 \]
Dividing by 5:
\[ x = \frac{2}{5} \]

### Step 7: Find \(y\)
Now substitute \(x = \frac{2}{5}\) back into Equation (1):
\[ \frac{2}{5} + y = 1 \]
So,
\[ y = 1 - \frac{2}{5} = \frac{3}{5} \]

### Conclusion
Thus, the fixed point through which the lines pass is:
\[ \left( \frac{2}{5}, \frac{3}{5} \right) \]