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Proof: (related to "Presentation of a Straight Line in a Plane as a Linear Equation")

By assumption, $L$ is a straight line in the number plane $\mathbb R^2.$

$L$ through the origin / homogenous equation

  • “$\Rightarrow$” Let $L$ go through the origin.
    • By a previous lemma, a parameterized description of $L$ using two vectors does not depend on a specific choice of these vectors.
    • Therefore, without loss of generality, we can choose the zero vector $\overrightarrow{OO}$ as the starting vector and an arbitrary point (but not the origin itself) $A\in L$ with the coordinates $A=(a_1,a_2)$ as a valid parameterized description of $L:=\{P\in\mathbb R^2:\;\overrightarrow{OP}=\overrightarrow{OO}+t\cdot \overrightarrow{OA}=t\cdot \overrightarrow{OA}=(t a_1,t a_2),\; t\in \mathbb R\}.$
    • Consider the set of points $S:=\{(x_1,x_2)\in\mathbb R^2:\;(x_1,x_2)\text{ solve }a_2x_1-a_1x_2=0.\}$
    • It follows that $L\subseteq S,$ since $a_2\cdot t a_1-a_1\cdot t a_2=0$ for all $t\in\mathbb R.$
    • Thus, $L$ is a solution of some homogenous equation.
  • “$\Leftarrow$” Let $L$ be the solution of a homogenous equation $\alpha_1x_1+\alpha_2x_2=0$.
    • To keep notation and the proof simple, we choose without loss of generality $\alpha_1:=a_2$ and $\alpha_2:=-a_1$ for some coordinates of a point $A=(a_1,a_2)$ which is not the origin.
    • Since $A$ was not the origin, not both $a_1,a_2$ can be zero.
    • Assume $a_1\neq 0.$ In this case a pair of numbers $(x_1,x_2)$ solves the equation $a_2x_1-a_1x_2=0$ if and only if $x_2=\frac {a_2}{a_1}x_1.$ This means that $S$ is a straight line with the parameterized description $S=\{Q\in\mathbb R^2:\;\overrightarrow{OQ}=(t,t\frac{a_2}{a_1}), t\in \mathbb R\}.$ This is a straight line going through the origin and the point $A=(a_1,a_2)$ (for $t=a_1$). Thus, by a previous lemma, $S=L$.
    • Now assume $a_1=0.$ In this case $a_2\neq 0$ and $a_2x_1=0$ holds only if $x_1=0$. Therefore, $S=\{Q\in\mathbb R^2:\;\overrightarrow{OQ}=(0,t a_2), t\in \mathbb R\}.$ Again, this is a straight line going through the origin and the point $A=(0,a_2)$ (for $t=1$). Again, previous lemma shows that we must have $S=L$.
    • Altogether, $L$ goes through the origin if and only if it equals the set $S$ of solutions of some homogenous linear equation $\alpha_1x_1+\alpha_2x_2=0$ (here with $\alpha_1:=a_2$ and $\alpha_2:=-a_1.$)

$L$ not through the origin / inhomogenous equation

  • “$\Rightarrow$” Let $L$ do not go through the origin.
    • Assume, that $L$ goes through two points $A=(a_1,a_2)$ and $B=(b_1,b_2)$ and that we have the parameterized description $L:=\{P\in\mathbb R^2:\;P:=(b_1,b_2) + (t a_1,t a_2),\; t\in \mathbb R\}.$
    • We have just proven that if $L_0:=\{Q\in\mathbb R^2:\;Q:=(t a_1,t a_2),\; t\in \mathbb R\}$, then $L_0$ is a straight line going through the origin and solving the homogenous linear equation $a_2x_1-a_1x_2=0.$
    • Therefore, $L=\{(b_1,b_2)+(t x_1,t x_2):\;a_2x_1-a_1x_2=0\}=\{(y_1,y_2):\; a_2y_1-a_1y_2=a_2b_1-a_1b_2\}.$
    • Since both, $A$ and $B$, are not origin, we have $\beta:=a_2b_1-a_1b_2\neq 0.$
    • Thus, $L$ is a solution of some inhomogenous equation.
  • “$\Leftarrow$” Let $L$ be the solution of an inhomogenous equation $\alpha_1x_1+\alpha_2x_2=\beta$ with $\beta\neq 0.$
    • We have seen that if $A=(a_1,a_2)$ is not the origin and lies on $L$ then we can find a straight line $L_0$ going through the origin and $A$ and solving the homogenous equation $a_2x_1-a_1x_2=0.$
    • Since $A$ lies also on $L$, we have by assumption $a_2b_1-a_1b_2=\beta$ for some point $B=(b_1,b_2)$.
    • Since $\beta\neq 0$ and $A$ is not the origin, we have that $B$ is also not the origin.
    • Moreover, $B\neq A$ (otherwise $a_2a_1-a_1a_2=0.$)
    • Thus, $L$ is a straight line going through two points $A$ and $B$, both not being the origin and the lines $L_0$ and $L$ meet at the point $A$.
    • Assume $L$ went through $A$ and the origin, like $L_0$ does. Then from the existence and uniqueness of a straight line through two points it would follow that $L$ and $L_0$ are identical, in contradiction with $\beta\neq 0$.
    • It follows that $L$ does not go through the origin.

| | | | created: 2018-10-12 13:32:34 | modified: 2018-10-12 20:24:50 | by: bookofproofs | references: [7937]

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Bibliography (further reading)

[7937] Knabner, P; Barth, W.: “Lineare Algebra – Grundlagen und Anwendungen”, Springer Spektrum, 2013

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