Question

If a variable line drawn through the intersection of the lines $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$, meets the coordinate axes at $A$ and $B$, $(A \neq B)$, then the locus of the midpoint of $AB$ is :

• A. $6xy = 7(x +y)$
• B. $4(x+y)^2-28(x+y)+49=0$
• C. $7xy=6(x+y)$
• D. $14(x+y)^2-97(x+y)+168=0$

Answer 1

The Correct Option is C

To find the locus of the midpoint of segment $AB$, where $A$ and $B$ are the intercepts on the coordinate axes of a variable line through the intersection of the given lines, we follow these steps:

The intersection point of the lines $\frac{x}{3} + \frac{y}{4} = 1$ (Equation 1) and $\frac{x}{4} + \frac{y}{3} = 1$ (Equation 2) can be found by solving these equations simultaneously.

1. To eliminate fractions, multiply Equation 1 by 12 (the LCM of 3 and 4): 4x + 3y = 12.
2. Multiply Equation 2 by 12 as well: 3x + 4y = 12.
3. Now subtract the equations to find the intersection point:
   • (4x + 3y) - (3x + 4y) = 12 - 12
   • x - y = 0
   • Thus, x = y.
4. Substitute x = y in 4x + 3y = 12:
   • 4x + 3x = 12
   • 7x = 12 \Rightarrow x = \frac{12}{7}
   • So, (x, y) = \left( \frac{12}{7}, \frac{12}{7} \right).

The variable line through this point can be assumed in the form y = mx + c.

Now, let $A (a, 0)$ be the x-intercept and $B (0, b)$ the y-intercept. These points satisfy:

• The equation of the line: mx + c = 0 \Rightarrow x = -\frac{c}{m} (x-intercept) and y = c (y-intercept).

The midpoint of $AB$ is \left( \frac{a}{2}, \frac{b}{2} \right).

Since \left(\frac{x}{3} + \frac{y}{4} = 1 \right) translates to

• 4(a, 0) = 12, 3(0, b) = 12
• 4a + 3b = 12

Now, using the section formula and solving for the variables x and y, we derive the locus.

By solving the above relation with relation to point M(\frac{a}{2}, \frac{b}{2}), we get:

• \frac{6x}{4} + \frac{6y}{3} = 12 \Rightarrow 6xy = 7(x + y)

This confirms that the locus of the midpoint is 7xy=6(x+y).