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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.

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  • $\begingroup$ May be $L_1 = A_1 x +B_1 \color{red}{ y} +C_1$? $\endgroup$ Commented May 14, 2016 at 12:46
  • $\begingroup$ copy paste error $\endgroup$ Commented May 14, 2016 at 12:48
  • $\begingroup$ This is not true. We can conclude, however, that $A_1 B_2 = A_2 B_1$. $\endgroup$ Commented May 14, 2016 at 13:10
  • $\begingroup$ @Travis desmos.com/calculator/dnc0szxvxb Look here it is true. $\endgroup$ Commented May 14, 2016 at 13:16
  • $\begingroup$ It should be L1: A1x + B1y +C1 = 0 rather than L1 = A1x+B1y+C1. $\endgroup$ Commented May 14, 2016 at 13:18

2 Answers 2

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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

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  • $\begingroup$ This solution seems to presuppose that $A_2, B_2$ are both nonzero. $\endgroup$ Commented May 14, 2016 at 13:09
  • $\begingroup$ Yes, these cases must be added separately $\endgroup$ Commented May 14, 2016 at 13:10
  • $\begingroup$ Sorry for the confusion but i mean in general form. that is we can divide the second equation by two and then $A_1$ = $A_2$ and $B_1$ = $B_2$ $\endgroup$ Commented May 14, 2016 at 13:11
  • $\begingroup$ Look here desmos.com/calculator/dnc0szxvxb $\endgroup$ Commented May 14, 2016 at 13:15
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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 $B_1=B_2$

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  • $\begingroup$ Let a line $A_1 x$ + $B_1 y $+ $C_1$ = 0 Now if you check in a graphing calculator then $A_1 x$ + $B_1 y $+ $C_2$ = 0 will be parallel to first line. So that means parallel lines have $x$ and $y$ - coefficient equal $\endgroup$ Commented May 14, 2016 at 12:58
  • $\begingroup$ @ritwiksinha Check the edits $\endgroup$ Commented May 14, 2016 at 13:03
  • $\begingroup$ look here, desmos.com/calculator/dnc0szxvxb This is what i mean. $\endgroup$ Commented May 14, 2016 at 13:15
  • $\begingroup$ @ritwiksinha What's the confusion? Dividing by $2$ is nothing but the same as $\frac{A_1}{A_2}=\frac{B_1}{B_2}=\frac{1}{2}$ $\endgroup$ Commented May 14, 2016 at 13:17
  • $\begingroup$ then you can see that $A_1$ = $A_2$ and same for the other. $\endgroup$ Commented May 14, 2016 at 13:30

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