Explanation

(c)
The equation of line passing through the point of intersection of $x + 2y - 1 = 0$ and $2x - y - 1 = 0$ is
$(x + 2y - 1) + \lambda(2x - y - 1) = 0$
$\Rightarrow x(1 + 2\lambda) + y(2 - \lambda) - 1 - \lambda = 0$
This meets the coordinate axes at $A\ \left( \frac{1 + \lambda)}{2\lambda + 1},\ 0 \right)\$and $B\left( 0,\frac{\lambda + 1}{2 - \lambda} \right)$
Let $(h,k)$ be the mid point of $AB$, then
$h = \frac{1}{2}\left( \frac{1 + \lambda}{2\lambda + 1} \right),\ k = \frac{1}{2}\left( \frac{\lambda + 1}{2 - \lambda} \right)$
On eliminating $\lambda$ from the these equations, we get
$h + 3k = 10hk$
Thus, the locus of $(h,\ k)$ is $x + 3y = 10xy$