Question

If a straight line drawn through the point of intersection of the lines \(4x + 3y - 1 = 0\) and \(3x + 4y - 1 = 0\), meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is:

• A. \(x + y - 7 = 0\)
• B. \(x + y - 14xy = 0\)
• C. \(2x + y + 14xy = 0\)
• D. \(x + 2y - 14xy = 0\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first find the point of intersection of the two given lines. Then, we write the equation of a general line passing through this point. By finding where this line intersects the $x$ and $y$ axes, we can determine the coordinates of $P$ and $Q$ and subsequently the locus of their midpoint $(h, k)$.
Step 2: Key Formula or Approach:
1. Intersection: Solve the system of linear equations.
2. Intercept form of a line: \(\frac{x}{a} + \frac{y}{b} = 1\), where \(P=(a, 0)\) and \(Q=(0, b)\).
3. Midpoint $(h, k)$: \(h = \frac{a}{2}\) and \(k = \frac{b}{2}\).
Step 3: Detailed Explanation:
Solving \(4x + 3y = 1\) and \(3x + 4y = 1\): Subtracting the equations gives \(x - y = 0 \implies x = y\). Substituting into the first: \(4x + 3x = 1 \implies 7x = 1 \implies x = 1/7, y = 1/7\). The point of intersection is \((1/7, 1/7)\). Let the line be \(\frac{x}{a} + \frac{y}{b} = 1\). Since it passes through \((1/7, 1/7)\): \[ \frac{1}{7a} + \frac{1}{7b} = 1 \implies \frac{1}{a} + \frac{1}{b} = 7 \] Let the midpoint be \((h, k)\). Then \(a = 2h\) and \(b = 2k\). Substituting these into the equation: \[ \frac{1}{2h} + \frac{1}{2k} = 7 \implies \frac{h+k}{2hk} = 7 \implies h + k = 14hk \] Replacing \((h, k)\) with \((x, y)\), we get \(x + y = 14xy\).
Step 4: Final Answer:
The locus of the midpoint is \(x + y - 14xy = 0\).