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 :

To find the locus of the midpoint of $$PQ$$, we first find the intersection point and then use the intercept form of a line.

Solve the system of equations:

1. $$4x + 3y = 1$$
2. $$3x + 4y = 1$$
3. $$P = (2h, 0)$$
4. $$Q = (0, 2k)$$

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

Substitute $$x = y$$ into (1): $$4x + 3x = 1 \implies 7x = 1 \implies x = \frac{1}{7}, y = \frac{1}{7}$$.

The point of intersection is $$A(\frac{1}{7}, \frac{1}{7})$$.

Let the midpoint of $$PQ$$ be $$M(h, k)$$.

Since $$P$$ lies on the $$x$$-axis and $$Q$$ lies on the $$y$$-axis, their coordinates are:

The intercept form of the line passing through $$P$$ and $$Q$$ is:

$$\frac{x}{2h} + \frac{y}{2k} = 1$$

Since this line passes through $$A(\frac{1}{7}, \frac{1}{7})$$, we substitute these values into the equation:

$$\frac{1/7}{2h} + \frac{1/7}{2k} = 1$$

$$\frac{1}{14h} + \frac{1}{14k} = 1$$

Multiply the entire equation by $$14hk$$ to clear the denominators:

$$k + h = 14hk$$

$$h + k - 14hk = 0$$

Replace $$(h, k)$$ with $$(x, y)$$ to get the final locus equation:

$$x + y - 14xy = 0$$

Correct Option: B