0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ $(ax+by+c)(a'x+b'y+c')=0 \iff \begin{cases}ax+by+c=0 \ \text{OR} \\a'x+b'y+c'=0\end{cases}$ ; 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
1
$\begingroup$

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.

enter image description here

See another answer here for your interest.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.