Question

The line passing through point of intersection of \(3x + 4y = 1\) and \(4x + 3y = 1\) intersects axes at P and Q, then locus of midpoint of PQ is

• A. \(\frac{1}{x} + \frac{1}{y} = 14\)
• B. \(\frac{3}{x} + \frac{4}{y} = 14\)
• C. \(\frac{4}{x} + \frac{3}{y} = 14\)
• D. \(x + y = 14\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
First, find the point of intersection of the two given lines. Any line passing through this point can be represented in the intercept form \(\frac{x}{a} + \frac{y}{b} = 1\). The midpoint of the intercepts \(P(a,0)\) and \(Q(0,b)\) is \((h, k) = (a/2, b/2)\). We then find the relationship between \(h\) and \(k\).

Step 2: Key Formula or Approach:
1. Solve \(3x + 4y = 1\) and \(4x + 3y = 1\). By symmetry or subtraction: \(x - y = 0 \implies x = y\). Substitute into the first: \(7x = 1 \implies x = 1/7, y = 1/7\). The intersection point is \((1/7, 1/7)\). 2. Line through \((1/7, 1/7)\) in intercept form: \(\frac{1/7}{a} + \frac{1/7}{b} = 1\).

Step 3: Detailed Explanation:
1. From the intercept form equation: \[ \frac{1}{7a} + \frac{1}{7b} = 1 \implies \frac{1}{a} + \frac{1}{b} = 7 \] 2. Let the midpoint of \(PQ\) be \((x, y)\). Then \(a = 2x\) and \(b = 2y\). 3. Substitute these into the equation: \[ \frac{1}{2x} + \frac{1}{2y} = 7 \] 4. Multiply by 2: \[ \frac{1}{x} + \frac{1}{y} = 14 \]

Step 4: Final Answer:
The locus of the midpoint is \(\frac{1}{x} + \frac{1}{y} = 14\).