If line `y - x - 1 +λ= 0` is equally inclined to axes and equidistant from the points`(1, - 2) and (3, 4)`, then `lambda` is

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the line given by the equation \( y - x - 1 + \lambda = 0 \) is equally inclined to the axes and equidistant from the points \( (1, -2) \) and \( (3, 4) \).

### Step 1: Understand the condition of being equally inclined to the axes
A line that is equally inclined to the axes has a slope of \( 1 \) or \( -1 \). Therefore, we can rewrite the line equation as:
\[ y - x + (\lambda - 1) = 0 \]
This implies that the slope of the line is \( 1 \) (since the coefficient of \( x \) is \( -1 \) and the coefficient of \( y \) is \( 1 \)).

### Step 2: Find the midpoint of the points \( (1, -2) \) and \( (3, 4) \)
The midpoint \( M \) of the points \( (1, -2) \) and \( (3, 4) \) is calculated as follows:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 3}{2}, \frac{-2 + 4}{2} \right) = \left( 2, 1 \right) \]

### Step 3: The line must pass through the midpoint
Since the line is equidistant from both points, it must pass through the midpoint \( (2, 1) \). We can substitute this point into the line equation:
\[ 1 - 2 - 1 + \lambda = 0 \]
This simplifies to:
\[ \lambda - 2 = 0 \]
Thus, we find:
\[ \lambda = 2 \]

### Step 4: Verify the conditions
1. The line equation becomes \( y - x - 1 + 2 = 0 \) or \( y - x + 1 = 0 \), which is indeed equally inclined to the axes (slope = 1).
2. The distance from the line \( y - x + 1 = 0 \) to the points \( (1, -2) \) and \( (3, 4) \) should be equal.

### Step 5: Calculate the distance from the line to the points
The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( y - x + 1 = 0 \), we have \( A = -1, B = 1, C = 1 \).

**Distance to point \( (1, -2) \)**:
\[ d_1 = \frac{|-1(1) + 1(-2) + 1|}{\sqrt{(-1)^2 + 1^2}} = \frac{|-1 - 2 + 1|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

**Distance to point \( (3, 4) \)**:
\[ d_2 = \frac{|-1(3) + 1(4) + 1|}{\sqrt{(-1)^2 + 1^2}} = \frac{|-3 + 4 + 1|}{\sqrt{2}} = \frac{|2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Since both distances are equal, the conditions are satisfied.

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = 2 \]