Why does the product of equations of straight lines represents 2 straight lines? Why does a 2nd degree equation is so linear?

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    $\begingroup$ $(ax+by+c)(a'x+b'y+c')=0 \iff \begin{cases}ax+by+c=0 \ \text{OR} \\a'x+b'y+c'=0\end{cases}$ ; the logical "OR" connective is very important because it is in connection whith the set connective "union", that's all : you have the union of two straight lines. $\endgroup$ – Jean Marie Feb 27 '18 at 19:37

In the language of conics,

\begin{align} 0 &= (ax+by+c)(a'x+b'y+c') \\[10pt] \iff 0 &= \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} aa' & \frac{ab'+a'b}{2} & \frac{ac'+a'c}{2} \\ \frac{ab'+a'b}{2} & bb' & \frac{bc'+b'c}{2} \\ \frac{ac'+a'c}{2} & \frac{bc'+b'c}{2} & cc' \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} \end{align}

Note that

$$\det \begin{pmatrix} aa' & \frac{ab'+a'b}{2} \\ \frac{ab'+a'b}{2} & bb' \end{pmatrix} =-\frac{(ab'-a'b)^2}{4}<0$$

providing non-parallel and

$$\det \begin{pmatrix} aa' & \frac{ab'+a'b}{2} & \frac{ac'+a'c}{2} \\ \frac{ab'+a'b}{2} & bb' & \frac{bc'+b'c}{2} \\ \frac{ac'+a'c}{2} & \frac{bc'+b'c}{2} & cc' \end{pmatrix} =0$$

implying a pair of straight lines.

It's a special case that the section plane cutting the radiant point of the cones.

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See another answer here for your interest.


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