If a straight line drawn through the point of intersection of the...

The point of intersection of the lines $4x+3y-1=0 and 3x+4y-1=0$ is obtained by solving them simultaneously.

Adding the two equations gives $7x+7y-2=0 \Rightarrow x+y=\dfrac{2}{7}$.

Subtracting the two equations gives $x-y=0 \Rightarrow x=y$.

Thus, the point of intersection is \left(\dfrac{1}{7}, \dfrac{1}{7}\right)$.<br><br>Let the equation of the line passing through this point and meeting the coordinate axes at$P and Q$be$\dfrac{x}{a} + \dfrac{y}{b} = 1$.<br><br>Since the line passes through \left(\dfrac{1}{7}, \dfrac{1}{7}\right)$, we have:

$\dfrac{1}{7a} + \dfrac{1}{7b} = 1 \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} = 7$

The coordinates of the points where the line meets the axes are $P(a, 0) and Q(0, b)$.

Let $(h, k)$ be the midpoint of $PQ$. Then $h = \dfrac{a}{2}$ and $k = \dfrac{b}{2}$, which gives $a = 2h and b = 2k$.

Substituting $a and b$ into the relation $\dfrac{1}{a} + \dfrac{1}{b} = 7$, we get:

$\dfrac{1}{2h} + \dfrac{1}{2k} = 7$

$\dfrac{h+k}{2hk} = 7$

$h+k = 14hk$

Replacing $(h, k)$ with $(x, y)$, the locus of the midpoint is:

$x+y-14xy=0$

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