Find the locus of the point of intersection of the lines `x+y=3+lamda and 5x-y=7+3lamda`, where `lamda` is a variable.

Answer

To find the locus of the point of intersection of the lines given by the equations \( x + y = 3 + \lambda \) and \( 5x - y = 7 + 3\lambda \), we will follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( x + y = 3 + \lambda \) (Equation 1)
2. \( 5x - y = 7 + 3\lambda \) (Equation 2)

### Step 2: Solve for \( x \) and \( y \)
To find the point of intersection, we can add the two equations to eliminate \( y \).

Adding Equation 1 and Equation 2:
\[ (x + y) + (5x - y) = (3 + \lambda) + (7 + 3\lambda) \]
This simplifies to:
\[ 6x = 10 + 4\lambda \]
Now, solving for \( x \):
\[ x = \frac{10 + 4\lambda}{6} = \frac{5 + 2\lambda}{3} \]

### Step 3: Substitute \( x \) back to find \( y \)
Now that we have \( x \), we can substitute it back into Equation 1 to find \( y \):
\[ y = 3 + \lambda - x \]
Substituting \( x \):
\[ y = 3 + \lambda - \frac{5 + 2\lambda}{3} \]
To simplify, we need a common denominator:
\[ y = \frac{9 + 3\lambda - (5 + 2\lambda)}{3} \]
This simplifies to:
\[ y = \frac{9 - 5 + 3\lambda - 2\lambda}{3} = \frac{4 + \lambda}{3} \]

### Step 4: Express \( \lambda \) in terms of \( x \) and \( y \)
From our expressions for \( x \) and \( y \):
1. \( x = \frac{5 + 2\lambda}{3} \)
2. \( y = \frac{4 + \lambda}{3} \)

We can express \( \lambda \) in terms of \( x \):
\[ 2\lambda = 3x - 5 \implies \lambda = \frac{3x - 5}{2} \]
And in terms of \( y \):
\[ \lambda = 3y - 4 \]

### Step 5: Set the two expressions for \( \lambda \) equal to each other
Since both expressions represent the same \( \lambda \):
\[ \frac{3x - 5}{2} = 3y - 4 \]

### Step 6: Clear the fraction and simplify
Multiply through by 2 to eliminate the fraction:
\[ 3x - 5 = 6y - 8 \]
Rearranging gives:
\[ 3x - 6y = -3 \]
Dividing through by 3:
\[ x - 2y = -1 \]

### Final Step: Rearranging to standard form
Rearranging gives us the locus equation:
\[ 2y - x = 1 \]

### Conclusion
The locus of the point of intersection of the lines is given by:
\[ \boxed{2y - x = 1} \]