A line segment of fixed length 2 units moves so that its ends are on the positive x-axis and on the part of the line$\mathrm{x}+\mathrm{y}=0$ which lies in the second quadrant. Then the locus of the midpoint of the line has the equations

As shown in the figure, line segment AB, of fixed length 2 units, slides such that one of its ends A is on x-axis and other ends A is x-axis and other end B is on line $\mathrm{x}+\mathrm{y}=0$

Let $\mathrm{P}\left(\mathrm{h},\mathrm{k}\right)$ be midpoint of AB.

Also, Let $\angle \mathrm{BAO}=\mathrm{\theta }.$

So, in triangle $\mathrm{AMB},\text{\hspace{0.17em}}\mathrm{BM}=2\mathrm{sin\theta }$ and $\mathrm{AM}=2\mathrm{cos\theta }.$

In triangle $\mathrm{BMO},\mathrm{MO}=\mathrm{MB}=2\mathrm{sin\theta }$

$\therefore A\equiv \left(2\cos \theta -2\sin \theta ,0\right)$ and $\mathrm{B}\equiv \left(-2\mathrm{sin\theta },2\mathrm{sin\theta }\right)$

Now, $\mathrm{P}\left(\mathrm{h},\mathrm{k}\right)$ is the midpoint of AB.

$\therefore 2\mathrm{h}=2\mathrm{cos\theta }-2\mathrm{sin\theta }+\left(-2\mathrm{sin\theta }\right)$ and $2\mathrm{k}=2\mathrm{sin\theta }$

$\Rightarrow \mathrm{h}=\mathrm{cos\theta }-2\mathrm{sin\theta }$ and $\mathrm{k}=\mathrm{sin\theta }$

Squaring and adding, we get 

${\left(\mathrm{h}+2\mathrm{k}\right)}^2+{\mathrm{k}}^2=1$ 

or ${\mathrm{x}}^2+5{\mathrm{y}}^2+4\mathrm{xy}-1=0$

This is the required locus.