4

Assume we have two equations of a line,

$A_1 x + B_1y = C_1$ and $A_2 x + B_2y = C_2$

Now we multiply the first equation by $B_2$ and the second by $B_1$ to obtain

(1) $A_1B_2x + B_1B_2y = C_1B_2$ and

(2) $A_2B_1x + B_1B_2y = C_2B_1$

Now, if we do (1) - (2) we get

$(A_1B_2 - A_2B_1)x = C_1B_2 - C_2B_1$

Then naturally we have,

$x = \frac{C_1B_2 - C_2B_1}{A_1B_2 - A_2B_1}$

Note that, $A_1B_2 - A_2B_1$ is the determinant of $(A_1, B_1)$ and $(A_2, B_2)$.

What does this mean? Why does the determinant come in the equation when trying to solve line-line intersections? What is the intuition behind dividing the equation by the determinant? Assume the determinant is non-zero.

Reference is here.

----------------Edit-------------

I understand that if the determinant is 0, they are parallel. My question is more of what does the determinant stand for in $x = \frac{C_1B_2 - C_2B_1}{A_1B_2 - A_2B_1}$ Is it an area, a vector length, etcetc. Why does the determinant end up in the denominator?

Community
  • 1
masotann
  • 201
  • 2
  • 6
  • If that determinant is zero then the two lines are either the same or parallel, so no unique point where they meet exists. – coffeemath Jun 19 '13 at 18:25
  • @coffeemath sorry let me add the assumption that the determinant is not 0. – masotann Jun 19 '13 at 18:27
  • 1
    If the determinant is not $0$ then the two lines meet, and $x$ is as you have it. Then $y$ can be got by using that $x$ value in either equation. Is your question about why the given $x$ formula is correct? Or is it about why the determinant ends up in the denominator? You might like to look up "Cramer's rule" about either of these questions. – coffeemath Jun 19 '13 at 18:49
  • @its about why the determinant ends up in the denominator. I'll take a peek at cramers rule then thanks! – masotann Jun 19 '13 at 18:50

2 Answers2

6

What does really an equation like $ax+by=c$ mean? Indeed it speaks about all $(x,y)\in\mathbb R^2$ such that satisfy that equation. Let's have another one, say $a'x+b'y=c'$. When we put these two equations together, in fact, we mean all $(x,y)\in\mathbb R^2$ such that satisffy two equations simultaneously. This word is what we have to find out. Simultaneously, means here to consider two equations and to solve them. This seems to us that we have a system of equations like:

$$ \left\{ \begin{array}{ll} ax+by=c \\ a'x+b'y=c' \end{array} \right. $$ There are several method in which you can solve this system. One is what you applied above. Obviously if we want two lines to be distinct and non parallel so we want them to satisfy: $$\frac{a}{b}\neq \frac{a'}{b'}$$ This is what you achieved above as determinant of $$\begin{pmatrix} a & b \\ a' & b'\\ \end{pmatrix}$$

Mikasa
  • 67,374
0

In the form $ax+by+c=0$, the pair $(a,b,0)$ represents a vector in the direction orthogonal to the line.

Given the two lines of normal vectors $\mathbf{n}_1=(a_1,b_1,0)$ and $\mathbf{n}_2=(a_2,b_2,0)$, we have $$ \mathbf{w}=\mathbf{n}_1\times \mathbf{n}_2=\left| \begin{align} \mathbf{i} && \mathbf{j} && \mathbf{k} \\ a_1 && b_1 && 0 \\ a_2 && b_2 && 0 \end{align} \right|=(a_1 b_2 - a_2 b_1)\mathbf{k} $$ Its length corresponds to the area of the parallelogram whose sides are $\mathbf{n}_1$ and $\mathbf{n}_2$, but this has no absolute meaning in relation to the problem, given that such vectors, like the equations of the lines, are defined up to a multiplicative constant.

Vincenzo Tibullo
  • 10,955