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Let $C$ be a cone generated by a line of gradient $m$ which goes through the origin. Then $(x,y,z)$ is on $C$ if $$(1) \qquad z = \pm m \sqrt{ x^{2} + y^{2} }.$$

Let $P$ be a plane which intersects with the cone and whose intersection with the $xy$-plane is a line parallel to the $y-$ plane. Thus, the intersection of $P$ with the $xz$-plane is a line: $L$, say. Supposing $L$ to have gradient $M$ and $z$-intercept $B$, the line $L$ can be described by the equation $$ (2) \qquad z = Mx+B.$$

All is right and well. But then he says 'combining $(1)$ and $(2)$, we see that (x,y,z) is in the intersection of the cone and the plane if and only if $$Mx+B = \pm m \sqrt{ x^{2} + y^{2} }.$$

I understand why, if $(x,y,z)$ is in the intersection, then $(1) = (2),$ but why, is the converse true? Surely we can find an infinite number of points where the equations are equal, but $z$ could be any number and the point not on either plane.

He doesn't first assume that the point is already on $C$ or $P,$ just that it is in $\mathbb{R}^{3},$ and I haven't missed anything in his argument out. Am I just being thick, or do I have a point?