Locus of Midpoint of Segment Through Intersection of Lines

To find the locus of the midpoint of the line segment $PQ$, we first determine the point of intersection of the two given lines: $4x + 3y - 1 = 0 \quad \text{--- (1)}$ $3x + 4y - 1 = 0 \quad \text{--- (2)}$

Adding equations (1) and (2): $7x + 7y - 2 = 0 \implies x + y = \frac{2}{7}$

Subtracting equation (2) from equation (1): $x - y = 0 \implies x = y$

Substituting $x = y$ into $x + y = \frac{2}{7}$, we get: $2x = \frac{2}{7} \implies x = \frac{1}{7} \quad \text{and} \quad y = \frac{1}{7}$

Thus, the point of intersection of the lines is $A\left(\frac{1}{7}, \frac{1}{7}\right)$.

Let $M(h, k)$ be the midpoint of the segment $PQ$. Since $P$ lies on the x-axis and $Q$ lies on the y-axis, their coordinates can be represented as: $P(2h, 0) \quad \text{and} \quad Q(0, 2k)$

The equation of the straight line passing through $P(2h, 0)$ and $Q(0, 2k)$ using the intercept form is: $\frac{x}{2h} + \frac{y}{2k} = 1$

Since this line passes through the point of intersection $A\left(\frac{1}{7}, \frac{1}{7}\right)$, substituting $x = \frac{1}{7}$ and $y = \frac{1}{7}$ gives: $\frac{1/7}{2h} + \frac{1/7}{2k} = 1$ $\frac{1}{14h} + \frac{1}{14k} = 1$

Multiplying both sides by $14hk$: $k + h = 14hk \implies h + k - 14hk = 0$

Replacing $(h, k)$ with $(x, y)$ to find the required locus: $x + y - 14xy = 0$

This matches Option B.