Explanation:
(B) : Given equation \(x+2 y=4\) and \(2 x+y=4\) meets the coordinate axes at \(\mathrm{A}\) and \(\mathrm{B}\).
\(x+2 y=4 \tag{i}\)
\(2 x+y=4 \tag{ii}\)
Solving equation (i) and (ii), we get-
\(x=\frac{4}{3}, y=\frac{4}{3} \tag{iii}\)
Let \(\quad \mathrm{AB}: \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
meet \(\mathrm{x}\)-axis at \(\mathrm{A}\) and \(\mathrm{y}\)-axis at \(\mathrm{B}\).
If straight line (iii) passes through the point \(\left(\frac{4}{3}, \frac{4}{3}\right)\) then
\(\frac{4}{3 \mathrm{a}}+\frac{4}{3 \mathrm{~b}}=1 \Rightarrow \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=\frac{3}{4} \tag{iv}\)
Let the mid-point of \(A B\) be \((h, k)\)
\(\text { So, } \mathrm{h}=\frac{\mathrm{a}+0}{2}, \mathrm{k}=\frac{0+\mathrm{b}}{2} \Rightarrow \mathrm{a}=2 \mathrm{~h}, \mathrm{~b}=2 \mathrm{k}\)
\(\therefore \quad \text { from equation (iv) }\)
\(\frac{1}{2 \mathrm{~h}}+\frac{1}{2 \mathrm{k}}=\frac{3}{4}\)
\(2 \mathrm{~h}+2 \mathrm{k}=3 \mathrm{hk}\)
replace \(\mathrm{x}, \mathrm{y}\) from \(\mathrm{h}\), \(\mathrm{k}\) respectively, we get-
\(2(x+y)=3 x y\)
\(\text { i.e, required locus. }\)
\(2(x+y)=3 x y\)