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: Finding the Point of Intersection:
Solve the system $3x + 4y = 1$ and $4x + 3y = 1$ simultaneously.
Subtracting the first from the second: $x - y = 0 \implies x = y$.
Substituting into $3x + 4x = 1 \implies 7x = 1 \implies x = \dfrac{1}{7}$.
So the point of intersection is $\left(\dfrac{1}{7}, \dfrac{1}{7}\right)$.
\includegraphics[width=0.5\linewidth]{m6.png}
Step 2: Setting up the Line through the Fixed Point:
Let the line intersect the $x$-axis at $P = (2h, 0)$ and the $y$-axis at $Q = (0, 2k)$, so that the midpoint of $PQ$ is $M = (h, k)$.
The equation of the line in intercept form is: \[ \frac{x}{2h} + \frac{y}{2k} = 1 \] Step 3: Applying the Condition that the Line Passes Through $\left(\dfrac{1}{7}, \dfrac{1}{7}\right)$:
\[ \frac{\frac{1}{7}}{2h} + \frac{\frac{1}{7}}{2k} = 1 \] \[ \frac{1}{14h} + \frac{1}{14k} = 1 \] \[ \frac{1}{h} + \frac{1}{k} = 14 \] Step 4: Writing the Locus:
Replacing $(h, k)$ by $(x, y)$: \[ \frac{1}{x} + \frac{1}{y} = 14 \] Step 5: Final Answer:
The locus of the midpoint of $PQ$ is $\dfrac{1}{x} + \dfrac{1}{y} = 14$.
The answer is Option (1).