A variable st.line drawn through the points of intersection of st.lines $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\text{ and }\frac{\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{y}}{\mathrm{a}}=1$ meets the co-ordinate axes at A and B. Show that the locus of the mid point of $\overline{\mathrm{AB}}$ is $2(\mathrm{a}+\mathrm{b})\mathrm{xy}=\mathrm{ab}(\mathrm{x}+\mathrm{y})$.  

$\text{ The st.lines }\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\text{ and }\frac{\mathrm{x}}{\mathrm{b}}+\frac{\mathrm{y}}{\mathrm{a}}=1\text{ intersect at }$$\mathrm{P}\text{ whose co-ordinates are }\left(\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}},\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\right)$
$\mathrm{Q}\left({\mathrm{x}}_0,{\mathrm{y}}_0\right)\text{ is a point on the given locus }$
$\text{ The st.line with }\mathrm{x}\text{-intercept ' }2{\mathrm{x}}_0\text{ ' }$
$\mathrm{y}\text{ - intercept }2{\mathrm{y}}_0\text{ passes through }\mathrm{p}$
$\begin{array}{l}\iff \mathrm{p}\text{ lies on st.line }\frac{\mathrm{x}}{2{\mathrm{x}}_0}+\frac{\mathrm{y}}{2{\mathrm{y}}_0}=1\\ \iff \left[\frac{1}{2{\mathrm{x}}_0}+\frac{1}{2{\mathrm{y}}_0}\right]\left(\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\right)=1\\ \iff 2(\mathrm{a}+\mathrm{b}){\mathrm{x}}_0{\mathrm{y}}_0=\mathrm{ab}\left({\mathrm{x}}_0+{\mathrm{y}}_0\right)\end{array}$
$\therefore \text{ the locus of }\mathrm{Q}\left({\mathrm{x}}_0{\mathrm{y}}_0\right)\text{ is }2(\mathrm{a}+\mathrm{b})\mathrm{xy}=\mathrm{ab}(\mathrm{x}+\mathrm{y})$