= Solution
The two affine lines meet at $(a,b)=(2,1)$. Translating their intersection to the origin, $X=x-2$, $Y=y-1$, removes both constant terms and gives the <homogeneous first-order differential equation>
$$
\frac{dY}{dX}=\frac{4X+2Y}{2X-Y}.
$$
On a patch with $X\ne0$, put $Y=uX$. Then $X\,du/dX=(u^2+4)/(2-u)$, and <separation of variables> gives
$$
\arctan(u/2)-\frac12\ln(u^2+4)=\ln|X|+C.
$$
Thus one local real implicit form is
$$
\boxed{\arctan\!\frac{y-1}{2(x-2)}-\frac12\ln\!\bigl(4(x-2)^2+(y-1)^2\bigr)=C.}
$$
The <arctangent> branch must be continued on the chosen solution patch. More geometrically, set $r=\sqrt{4X^2+Y^2}$ and take a continuous local argument $\theta=\arg(2X+iY)$; then $\theta-\ln r=C$. This form extends through $X=0$ where the original slope is defined. The intersection itself is excluded because the original right-hand side is $0/0$ there, and points with $2X-Y=0$ do not define a finite derivative. No real constant-$u$ solution was lost, because $u^2+4$ has no real zero. Translating the intersection works precisely because the two affine linear forms then become homogeneous linear forms.
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