0

Let two lines to be parallel in their general form.

$L_1$ : $A_1 x$ + $B_1 y$ + $C_1$

$L_2$ : $A_2 x$ + $B_2 y$ + $C_2$

Now i wish to prove $A_1$ = $A_2$ and $B_1$ = $B_2$

But i can only think of the prove in my head, not sure how to write it mathematically.

  • May be $L_1 = A_1 x +B_1 \color{red}{ y} +C_1$? – Roman83 May 14 '16 at 12:46
  • copy paste error –  May 14 '16 at 12:48
  • This is not true. We can conclude, however, that $A_1 B_2 = A_2 B_1$. – Travis Willse May 14 '16 at 13:10
  • @Travis https://www.desmos.com/calculator/dnc0szxvxb Look here it is true. –  May 14 '16 at 13:16
  • It should be L1: A1x + B1y +C1 = 0 rather than L1 = A1x+B1y+C1. – Zau May 14 '16 at 13:18
  • Ok i will edit it. Sorry i dont know so minute details. –  May 14 '16 at 13:20
  • That link shows the opposite: In fact, it demonstrates that given a line $A x + B y + C = 0$, the equation $(\lambda A) x + (\lambda B) y + (\lambda C) = 0$, where $\lambda \neq 0$, defines the same line. If we take $\lambda \neq 1$, then the coefficients of $x$ in the two equations are different, contradicting your claim. – Travis Willse May 14 '16 at 13:20
  • Now three people have showed you that your claim is not true, and yet you keep saying it is and sending "proof" of this claim, which is just one example in which your claim holds. Can you accept that your claim is not true? – B. Pasternak May 14 '16 at 13:21
  • @B.Pasternak how ? Look if we take two equal lines Say $A_1 x$ + $B_1 x$ = 0, we can offset one of the lines by any amount $C_1$ and then the lines will be parallel. That is now line $A_1 x$ + $B_1 x$ = 0 and $A_1 x$ + $B_1 x$ + $C_1$ = 0 are parallel. –  May 14 '16 at 13:25
  • Do you understand that the implication (assuming $A_2\neq0$ and $B_2\neq0$) $\frac{A_1}{A_2}=\frac{B_1}{B_2}\implies A_1=A_2\wedge B_1=B_2$ does not hold in general (consider $\frac{1}{2}=\frac{2}{4}$), but the implication $A_1=A_2\wedge B_1=B_2\implies\frac{A_1}{A_2}=\frac{B_1}{B_2}$ (assuming $A_2\neq0$ and $B_2\neq0$) does hold in general? – B. Pasternak May 14 '16 at 13:29
  • Ok, let two lines be parallel, $A_1 x$ + $B_1 y$ + $C_1$ = 0 and $A_2 x$ + $B_2 y$ + $C_2$ = 0, now the second line can be written in form $A_1 x$ + $B_1 y$ + $\frac{C_2}{a}$ = 0, where $a$ is any real number. Do you agree with this ? –  May 14 '16 at 13:36

2 Answers2

1

Let $A_1, A_2, B_1,B_2 \not = 0$. Then

$$L_1 || L_2 \Leftrightarrow \frac {A_1}{A_2}=\frac {B_1}{B_2}$$ Proof:

Let $\frac {A_1}{A_2}\not=\frac {B_1}{B_2}$. Then the system of equations $$\begin{cases} A_1 x + B_1 y + C_1=0, \\ A_2 x + B_2 y + C_2=0 \end{cases} $$

has a solution. A point that belongs to both straight. The contradiction (because the lines are parallel)

Addition: For example:

$l_1: 2x+3y+7=0$ and $4x+6y+7=0$

enter image description here

Roman83
  • 17,884
  • 3
  • 26
  • 70
0

Your claim is wrong.

Write the formula of each line in the form $y=mx +c$

$L_1: y= (\frac{-A_1}{B_1})x + (\frac{-C_1}{B_1})$

$L_2: y= (\frac{-A_2}{B_2})x + (\frac{-C_2}{B_2})$

For two lines to parallel theirs slopes($m$) must be equal.

So, by comparison, it would be seen that $\frac{A_1}{B_1}=\frac{A_2}{B_2}$,

which doesn't necessarily $\Rightarrow A_1=A_2$ and $B_1=B_2$

Roby5
  • 4,287