Problem:
A rod of length $2$ units moves so that its ends are on the positive $x$-axis and on the line $x+y=0$ which lies in the second quadrant. Find the locus of the midpoint of the rod.
My approach:
Let the midpoint be $P(h,k).$ Let the end on the $x$-axis have coordinates $(x_2,0$). The end on $x+y=0$ has coordinates $(x_1,y_1)$. $$\Rightarrow 2h=x_1+x_2$$ $$\Rightarrow 2k=y_1$$ $$\Longrightarrow 2h+2k=x_2$$
I cannot eliminate $x_2$ from the equation.
On
If we let parameter $t$ vary from 0 to 2, following $x$-coordinate of the right end of the rod $R_t=(t,0)$, then the left end has coordinates $L_t=(-s,s)$, where $s$ can be found from
\begin{align} 2^2&=s^2+(t+s)^2, \\ s&=\tfrac12(\sqrt{8-t^2}-t) \end{align} Thus coordinates of the midpoint $M$ in terms of $t$ are \begin{align} M_x&=\tfrac14(3 t-\sqrt{8-t^2}), \\ M_y&=\tfrac14(-t+\sqrt{8-t^2}). \end{align}
We can also exclude $t$ and get \begin{align} y(x)&=\tfrac15\left(\sqrt{5-x^2}-2x\right),\quad x\in[-\tfrac{\sqrt2}2,1]. \end{align}
You have not used the information of the length of the rod yet. That gives you $$(x_2-x_1)^2+(0-y_1)^2=4$$
If you plug $x_2=2h+2k, x_1=-2k , y_1=2k$ into this, you will find an equation about $h,k$.