= Solution
At a <Lagrange point>, the test particle is stationary in the <rotating reference frame>, so $F=G=0$. On the $x$ axis, $y=0$ makes $G=0$ automatically. The singularities at the two masses split the axis into three intervals, and the balance between gravity and <centrifugal acceleration> gives one root of $F=0$ in each interval. These are the three <Collinear Lagrange points> $L_1,L_2,L_3$.
For an off-axis equilibrium, $y\ne0$. Put $A_i=1-r_i^{-3}$. Then
$$
\frac Gy=\mu_1A_1+\mu_2A_2=0,
$$
while
$$
F-x\frac Gy=\mu_1\mu_2(A_1-A_2)=0.
$$
Hence $A_1=A_2=0$ and therefore $r_1=r_2=1$. The two intersections of unit circles centered at $M_1$ and $M_2$ form equilateral triangles with the binary, giving the <Triangular Lagrange points>
$$
\boxed{x=\frac{\mu_1-\mu_2}{2},qquad y=\pm\frac{\sqrt3}{2}}.
$$
Together with the three collinear points, these are the five equilibria of the <circular restricted three-body problem>.
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