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Statement. For a planar point $p$ and an oriented line $\lambda$, one regulus has affine rulings

$$A=\{\ell_{p,a}:a\in\lambda\},\qquad B=\{S(\lambda',\lambda): p\in\lambda',\ v_{\lambda'}\ne v_\lambda\}. \tag{1}$$

Here directions are unit oriented directions; opposite directions are allowed. Every line in $B$ is horizontal. Reflecting $z\mapsto-z$ gives the version with $\ell_{a,p}$ and $S(\lambda,\lambda')$.

Source and correction. Mathialagan, published 2021 PDF, pp. 21--23, Proposition 42. On p. 22 the sentence asserting that no nonhorizontal line is “in $R$” follows an argument about transversals. Its valid scope is the opposite/transversal ruling; the whole $R$ certainly contains the nonhorizontal generating lines. The coordinate proof below makes this distinction explicit and uses the corrected sign convention of Proposition 27.

Normalizing coordinates. A simultaneous orientation-preserving planar isometry $t\mapsto Ut+v$ sends rotation centers to $Uo+v$ and leaves the angle, hence $z$, unchanged. Equation (1) of Proposition 27 is equivariant under it, since $UJ=JU$. We may therefore take $p=(0,0)$ and $\lambda=\{(a,b):a\in\mathbb R\}$, with direction $(1,0)$ and some fixed $b\in\mathbb R$.

The lines of $A$ are then

$$(x,y,z)=\left(\frac{a+bz}{2},\frac{b-az}{2},z\right), \qquad z\in\mathbb R. \tag{2}$$

Their union is exactly the quadratic surface

$$F(x,y,z)=2y-b+2xz-bz^2=0. \tag{3}$$

Indeed, solving (3) at fixed $z$ gives $y=(b-2xz+bz^2)/2$; put $a=2x-bz$ to recover (2). The homogeneous quadratic is $2YW-bW^2+2XZ-bZ^2$. Its gradient vanishes only at the zero vector: the $Y$ derivative forces $W=0$, the $X$ derivative forces $Z=0$, then the $Z$ and $W$ derivatives force $X=Y=0$. It is therefore a smooth projective quadric. Any three distinct lines (2) are pairwise skew, so the uniqueness in Proposition 36 identifies (3) as their regulus.

Classifying every affine line. At each fixed height $z=t$, (3) is one horizontal line $H_t$. A nonhorizontal line can be written $x=A+Bz$, $y=C+Dz$. Substitution into (3) gives the three coefficient equations

$$2C-b=0,\qquad 2D+2A=0,\qquad 2B-b=0.$$

Hence it is precisely (2) with $a=2A$. There are no other nonhorizontal lines, and every horizontal line contained in the surface must equal its full slice $H_t$. Lines (2) form one ruling, while the $H_t$ form the other: each $H_t$ meets each line (2), and distinct lines within either family are projectively disjoint by the ruling description of Proposition 36.

The interpretation of the horizontal lines. For each $t$ take $\theta$ with $t=-\cot(\theta/2)$ and let $\lambda'$ be the line through $p$ of oriented direction $R_\theta^{-1}(1,0)$. A rotation of this angle takes $\lambda'$ onto $\lambda$ precisely when it takes $p$ onto $\lambda$. Formula (2) shows that the centers of all such rotations are exactly $H_t$. By Proposition 28 this is $S(\lambda',\lambda)$. Conversely, every $\lambda'$ in (1) fixes a unique nonzero angle and therefore gives one such $H_t$. This proves (1), including all horizontal lines of the opposite ruling. Reflection interchanges a motion and its inverse, proving the symmetric statement.

Dependencies and verification. Verified within the independently reviewed Theorem 3 chain, retained in the final review; Propositions 27, 28 and 36 supply the conventions and geometry. The explicit coefficient proof replaces the source's exhaustion argument and avoids the external Lemma 39. It is part of the living Theorem 3 record.

Bears on. Problem 661.