Locus of point of intersection of the perpendicular lines one belonging to `(x + y -2) + lamda (2x+3y-5)=0` and other to `(2x+y -11) + lamda(x+2y-13)=0` is a

Options

A. circle
B. Straight line
C. pair of lines
D. None of these

Answer

To find the locus of the point of intersection of the perpendicular lines given by the equations:

1. \((x + y - 2) + \lambda(2x + 3y - 5) = 0\)
2. \((2x + y - 11) + \lambda(x + 2y - 13) = 0\)

we will follow these steps:

### Step 1: Identify the Family of Lines
The first equation can be expressed as:
\[ x + y - 2 + \lambda(2x + 3y - 5) = 0 \]
This represents a family of lines parameterized by \(\lambda\).

The second equation can be expressed as:
\[ 2x + y - 11 + \lambda(x + 2y - 13) = 0 \]
This also represents another family of lines parameterized by \(\lambda\).

### Step 2: Find the Intersection Points
To find the intersection points of these lines, we need to solve the two equations for specific values of \(\lambda\).

**For the first family of lines:**
1. Set \(\lambda = 0\):
\[ x + y - 2 = 0 \quad \text{(Line 1)} \]
2. Set \(\lambda = 1\):
\[ 2x + 3y - 5 = 0 \quad \text{(Line 2)} \]

Now, solve these two equations:
1. From Line 1: \(y = 2 - x\)
2. Substitute \(y\) in Line 2:
\[ 2x + 3(2 - x) - 5 = 0 \implies 2x + 6 - 3x - 5 = 0 \implies -x + 1 = 0 \implies x = 1 \]
Substitute \(x = 1\) into Line 1:
\[ y = 2 - 1 = 1 \]
So, the intersection point is \((1, 1)\).

**For the second family of lines:**
1. Set \(\lambda = 0\):
\[ 2x + y - 11 = 0 \quad \text{(Line 3)} \]
2. Set \(\lambda = 1\):
\[ x + 2y - 13 = 0 \quad \text{(Line 4)} \]

Now, solve these two equations:
1. From Line 3: \(y = 11 - 2x\)
2. Substitute \(y\) in Line 4:
\[ x + 2(11 - 2x) - 13 = 0 \implies x + 22 - 4x - 13 = 0 \implies -3x + 9 = 0 \implies x = 3 \]
Substitute \(x = 3\) into Line 3:
\[ y = 11 - 2(3) = 5 \]
So, the intersection point is \((3, 5)\).

### Step 3: Set Up the Perpendicular Condition
Let the intersection point of the two families of lines be \((h, k)\). The slopes of the lines can be derived from the equations:
- For the first family, the slope from \(x + y - 2 = 0\) is \(-1\) and from \(2x + 3y - 5 = 0\) is \(-\frac{2}{3}\).
- For the second family, the slope from \(2x + y - 11 = 0\) is \(-2\) and from \(x + 2y - 13 = 0\) is \(-\frac{1}{2}\).

Since the lines are perpendicular, the product of their slopes must equal \(-1\):
\[ \left(\frac{k - 1}{h - 1}\right) \left(\frac{k - 5}{h - 3}\right) = -1 \]

### Step 4: Simplify the Equation
Cross-multiplying gives:
\[ (k - 1)(k - 5) = -(h - 1)(h - 3) \]
Expanding both sides:
\[ k^2 - 6k + 5 = -h^2 + 4h - 3 \]
Rearranging gives:
\[ h^2 + k^2 - 4h - 6k + 8 = 0 \]

### Step 5: Identify the Locus
This equation represents a circle. The locus of the point of intersection of the perpendicular lines is given by:
\[ h^2 + k^2 - 4h - 6k + 8 = 0 \]

### Final Result
The locus of the point of intersection is:
\[ x^2 + y^2 - 4x - 6y + 8 = 0 \]