Key Concepts and Formulas

• Pencil of Lines: The equation of any line passing through the intersection of two lines $L_1 = 0$ and $L_2 = 0$ is given by $L_1 + \lambda L_2 = 0$, where $\lambda$ is a real parameter.
• Intercepts: The x-intercept is the point where $y=0$, and the y-intercept is the point where $x=0$.
• Midpoint Formula: The midpoint of two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$.

Step-by-Step Solution

Step 1: Convert the given equations to the general form

The given lines are $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$. We convert these to the form $Ax + By + C = 0$.

• Line 1: $\frac{x}{3} + \frac{y}{4} = 1$. Multiplying by 12, we get $4x + 3y = 12$, so $L_1: 4x + 3y - 12 = 0$.
• Line 2: $\frac{x}{4} + \frac{y}{3} = 1$. Multiplying by 12, we get $3x + 4y = 12$, so $L_2: 3x + 4y - 12 = 0$.

We do this to easily apply the pencil of lines formula.

Step 2: Write the equation of the variable line

The variable line passes through the intersection of $L_1 = 0$ and $L_2 = 0$. Its equation is given by $L_1 + \lambda L_2 = 0$. Substituting the expressions for $L_1$ and $L_2$, we have: $(4x + 3y - 12) + \lambda (3x + 4y - 12) = 0$ $4x + 3y - 12 + 3\lambda x + 4\lambda y - 12\lambda = 0$ $x(4 + 3\lambda) + y(3 + 4\lambda) - 12(1 + \lambda) = 0$ This represents the equation of any line passing through the intersection of the two given lines.

Step 3: Find the x and y intercepts

The variable line meets the coordinate axes at A and B.

• x-intercept (Point A): Set $y = 0$ in the equation of the line: $x(4 + 3\lambda) - 12(1 + \lambda) = 0$ $x = \frac{12(1 + \lambda)}{4 + 3\lambda}$ So, point A is $\left( \frac{12(1 + \lambda)}{4 + 3\lambda}, 0 \right)$.

• y-intercept (Point B): Set $x = 0$ in the equation of the line: $y(3 + 4\lambda) - 12(1 + \lambda) = 0$ $y = \frac{12(1 + \lambda)}{3 + 4\lambda}$ So, point B is $\left( 0, \frac{12(1 + \lambda)}{3 + 4\lambda} \right)$.

Step 4: Find the midpoint of AB

Let $(h, k)$ be the midpoint of AB. Using the midpoint formula: $h = \frac{\frac{12(1 + \lambda)}{4 + 3\lambda} + 0}{2} = \frac{6(1 + \lambda)}{4 + 3\lambda} \quad \text{(Equation 1)}$ $k = \frac{0 + \frac{12(1 + \lambda)}{3 + 4\lambda}}{2} = \frac{6(1 + \lambda)}{3 + 4\lambda} \quad \text{(Equation 2)}$

Step 5: Eliminate the parameter λ

From Equation 1: $h(4 + 3\lambda) = 6(1 + \lambda)$ $4h + 3h\lambda = 6 + 6\lambda$ $3h\lambda - 6\lambda = 6 - 4h$ $\lambda(3h - 6) = 6 - 4h$ $\lambda = \frac{6 - 4h}{3h - 6} = \frac{2(3 - 2h)}{3(h - 2)} \quad \text{(Equation 3)}$

From Equation 2: $k(3 + 4\lambda) = 6(1 + \lambda)$ $3k + 4k\lambda = 6 + 6\lambda$ $4k\lambda - 6\lambda = 6 - 3k$ $\lambda(4k - 6) = 6 - 3k$ $\lambda = \frac{6 - 3k}{4k - 6} = \frac{3(2 - k)}{2(2k - 3)} \quad \text{(Equation 4)}$

Equate Equation 3 and Equation 4: $\frac{2(3 - 2h)}{3(h - 2)} = \frac{3(2 - k)}{2(2k - 3)}$ $4(3 - 2h)(2k - 3) = 9(h - 2)(2 - k)$ $4(6k - 9 - 4hk + 6h) = 9(2h - hk - 4 + 2k)$ $24k - 36 - 16hk + 24h = 18h - 9hk - 36 + 18k$ $6h + 6k - 7hk = 0$ $6(h + k) = 7hk$

Step 6: Write the locus equation

Replace $h$ with $x$ and $k$ with $y$: $6(x + y) = 7xy$ $7xy = 6(x + y)$

Common Mistakes & Tips

• Be careful with signs when expanding and simplifying the equations. A small sign error can lead to an incorrect locus.
• Always check for restrictions on the parameter $\lambda$. In this case, while not strictly necessary for solving, $\lambda \ne -1$ because it would imply that the line passes through the origin, which is not allowed due to the condition $A \ne B$. Also, $\lambda \ne -4/3$ and $\lambda \ne -3/4$.
• Remember to replace $h$ and $k$ with $x$ and $y$ at the end to get the final locus equation.

Summary

We found the equations of the two lines in the standard form, then used the pencil of lines concept to obtain the equation of a general line passing through their intersection. We then found the x and y intercepts of this line, and subsequently the midpoint of the segment joining them. Finally, we eliminated the parameter $\lambda$ to find the locus of the midpoint, which is $7xy = 6(x + y)$.

The final answer is $\boxed{\text{6xy = 7(x + y)}}$, which corresponds to option (A).