Consider the family of line `(x+y-1)+lamda(2x+3y-5)=0` and `(3x+2y-4)+mu(x+2y-6)=0` Equation of a straight line that belongs to both the families is:

Options

A. `x-2y-8=0`
B. `x-2y+8=0`
C. `2x-y-8=0`
D. None of these

Answer

To find the equation of a straight line that belongs to both families given by the equations:

1. \((x + y - 1) + \lambda(2x + 3y - 5) = 0\)
2. \((3x + 2y - 4) + \mu(x + 2y - 6) = 0\)

we will follow these steps:

### Step 1: Rewrite the equations
First, we can rewrite the equations in a more standard form.

For the first family:
\[ x + y - 1 + \lambda(2x + 3y - 5) = 0 \]
This can be rearranged as:
\[ (1 + 2\lambda)x + (1 + 3\lambda)y - (1 + 5\lambda) = 0 \]

For the second family:
\[ 3x + 2y - 4 + \mu(x + 2y - 6) = 0 \]
This can be rearranged as:
\[ (3 + \mu)x + (2 + 2\mu)y - (4 + 6\mu) = 0 \]

### Step 2: Set the coefficients equal
For the two lines to be the same, their coefficients must be proportional. Therefore, we set up the following equations based on the coefficients of \(x\), \(y\), and the constant term:

1. \(\frac{1 + 2\lambda}{3 + \mu} = \frac{1 + 3\lambda}{2 + 2\mu} = \frac{-(1 + 5\lambda)}{-(4 + 6\mu)}\)

### Step 3: Solve the proportional equations
Let's solve the first two ratios:
\[ \frac{1 + 2\lambda}{3 + \mu} = \frac{1 + 3\lambda}{2 + 2\mu} \]

Cross-multiplying gives:
\[ (1 + 2\lambda)(2 + 2\mu) = (1 + 3\lambda)(3 + \mu) \]

Expanding both sides:
\[ 2 + 2\mu + 4\lambda + 4\lambda\mu = 3 + \mu + 9\lambda + 3\lambda\mu \]

Rearranging gives:
\[ 4\lambda\mu - 3\lambda\mu + 2\mu - \mu + 4\lambda - 9\lambda + 2 - 3 = 0 \]
\[ (4\lambda - 9\lambda + 2\mu - \mu - 1) + (4\lambda\mu - 3\lambda\mu) = 0 \]

This simplifies to:
\[ -5\lambda + \mu - 1 + (4\lambda - 3)\mu = 0 \]

### Step 4: Solve for \(\lambda\) and \(\mu\)
Now we can express \(\mu\) in terms of \(\lambda\):
\[ \mu = 5\lambda + 1 \]

### Step 5: Substitute back
Now we can substitute \(\mu\) back into one of the original equations to find a specific line. We can use the second family of lines:
\[ (3 + (5\lambda + 1))x + (2 + 2(5\lambda + 1))y - (4 + 6(5\lambda + 1)) = 0 \]

### Step 6: Simplify the equation
This gives:
\[ (4 + 5\lambda)x + (7 + 10\lambda)y - (34 + 30\lambda) = 0 \]

### Step 7: Choose a specific value for \(\lambda\)
To find a specific line, we can choose \(\lambda = 0\):
\[ 4x + 7y - 34 = 0 \]

### Final Equation
Thus, the equation of the straight line that belongs to both families is:
\[ 4x + 7y - 34 = 0 \]