Detailed Solution

This problem asks for the locus of the mid-point of a line segment formed by the intercepts of a variable line on the coordinate axes. This variable line has a specific property: it must pass through the point where two other fixed lines intersect.

Intersection Point Calculation

First, we need to find the point where the two given lines, \(x + 2y - 1 = 0\) and \(2x - y - 1 = 0\), intersect. We can solve this system of linear equations.

• The given equations are:
  \(x + 2y - 1 = 0\) (Equation 1)
  \(2x - y - 1 = 0\) (Equation 2)
• To eliminate \(y\), we can multiply Equation 2 by 2:
  \(2 \times (2x - y - 1) = 0 \implies 4x - 2y - 2 = 0\) (Equation 3)
• Now, add Equation 1 and Equation 3:
  \((x + 2y - 1) + (4x - 2y - 2) = 0\)
  \(5x - 3 = 0\)
  \(x = \frac{3}{5}\)
• Substitute the value \(x = \frac{3}{5}\) back into Equation 1:
  \(\frac{3}{5} + 2y - 1 = 0\)
  \(2y = 1 - \frac{3}{5}\)
  \(2y = \frac{2}{5}\)
  \(y = \frac{1}{5}\)
• Thus, the point of intersection is \(P\left(\frac{3}{5}, \frac{1}{5}\right)\).

Variable Line Intercepts Analysis

Consider a variable line passing through the intersection point \(P\left(\frac{3}{5}, \frac{1}{5}\right)\). Let its equation in the intercept form be:

This line intersects the x-axis at \(A(a, 0)\) and the y-axis at \(B(0, b)\).

Since the line passes through \(P\left(\frac{3}{5}, \frac{1}{5}\right)\), these coordinates must satisfy the line's equation:

Multiply by \(5ab\) to clear the denominators:

This equation establishes a relationship between the intercepts \(a\) and \(b\). Let's call this (Equation *).

Mid-point Coordinates Derivation

We are interested in the mid-point of the line segment \(AB\), where \(A = (a, 0)\) and \(B = (0, b)\). Let the coordinates of this mid-point be \((h, k)\).

Using the mid-point formula:

This expresses the intercepts \(a\) and \(b\) in terms of the mid-point coordinates \(h\) and \(k\).

Locus Equation Determination

Now, substitute \(a = 2h\) and \(b = 2k\) into Equation (*), which is \(a + 3b = 5ab\):

Divide the entire equation by 2 to simplify:

This equation relates the coordinates \((h, k)\) of the mid-point. To find the locus, we replace \(h\) with \(x\) and \(k\) with \(y\).

The locus of the mid-point of \(AB\) is therefore: