Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to show that any line parallel to $Ax+ By + c =0$ is of the form $Ax + By + k =0$

How do I show this?

Thank you!

share|improve this question

4 Answers 4

up vote 2 down vote accepted

(Based on zar's answer and Moron's comment there.)

If two lines $Ax+By+C=0$ and $Dx+Ey+F=0$ are parallel, then they have no points of intersection (or infinitely many if you allow a line to be parallel to itself). In either case, the system of equations $$\begin{cases} Ax+By+C=0,\\ Dx+Ey+F=0 \end{cases}$$ will not have a single, unique solution. The system of equations is equivalent to $$\begin{bmatrix} A & B\\ D & E \end{bmatrix}\cdot\begin{bmatrix} x\\ y \end{bmatrix}+\begin{bmatrix} C\\ F \end{bmatrix}=0$$ or $$\begin{bmatrix} A & B\\ D & E \end{bmatrix}\cdot\begin{bmatrix} x\\ y \end{bmatrix}=\begin{bmatrix} -C\\ -F \end{bmatrix}.$$ This matrix equation has a single unique solution if and only if the coefficient matrix $\begin{bmatrix} A & B\\ D & E \end{bmatrix}$ is invertible, which is true if and only if its determinant is nonzero.

So, if the lines are parallel, then $$\begin{vmatrix} A & B\\ D & E \end{vmatrix}=AE-BD=0$$ or $$\frac{A}{D}=\frac{B}{E}$$ which means that $\frac{A}{D}(Dx+Ey+F)=Ax+By+\frac{FA}{D}=0$, so $Dx+Ey+F=0$ can be written in the form $Ax+By+k=0$.

edit: from Arturo Magidin's comment below, to handle the cases where $D=0$ or $E=0$ (in which $A/D = B/E$ makes no sense):

If $D=0$ then $A=0$ or $E=0$, but you cannot have both $D$ and $E$ zero. So $D=0$ implies $A=0$, and $B\neq 0\neq E$. Similarly, if $E=0$, then $B=0$ and $A\neq 0 \neq D$. And you get the similar conclusion without forgetting about horizontal and vertical lines.

share|improve this answer
2  
You also need to allow for the possibility that $D=0$ or $E=0$, so that $A/D = B/E$ make no sense. If $D=0$ then $A=0$ or $E=0$, but you canno thave both $D$ and $E$ zero. So $D=0$ implies $A=0$, and $B\neq 0\neq E$. Similarly, if $E=0$, then $B=0$ and $A\neq 0 \neq D$. And you get the similar conclusion without forgetting about horizontal and vertical lines. –  Arturo Magidin Sep 18 '10 at 20:40
    
@Arturo: Ahh, yes, good point. –  Isaac Sep 18 '10 at 20:49

Examples don't show it in any case. You cannot prove that something holds for infinitely many things by doing nothing except checking finitely many examples.

What you need to do is remember that two lines are parallel if and only if they have the same slope or they are both vertical (of the form $x=k$ for some constant $k$). Then figure out what the slope of an arbitrary line $\alpha x+\beta y+\gamma=0$ would be in terms of $\alpha$, $\beta$, and $\gamma$ (careful when dividing by constants to consider what happens if the constant you are dividing by is equal to zero). Then see what this means if you want to find all lines $\alpha x+\beta y+\gamma = 0$ that are parallel to a given line $Ax+By+c=0$.

share|improve this answer
    
I think "You cannot prove that something holds for infinitely many things by checking finitely many examples" is too strong a statement to be true. Certainly though, you can't prove this theorem by picking a set of 20 different values of A,B,c,k and checking. –  anon Sep 18 '10 at 19:40
    
Fair enough; should be "only by checking finitely many examples". You could do it if you also reduce the problem to a finite set of potential counterexamples in addition to checking those exemples. –  Arturo Magidin Sep 18 '10 at 19:46

It is a preliminary and useful exercise to show that the line $Ax+By+C=0$ and the vector $v=(A,B)$ are perpendicular (one says that $v$ is the normal vector of the line). With these, all amounts to observe that two lines are parallel if and only if their normal vectors are parallel.

share|improve this answer

You can solve the following system of equations:

$$ \begin{cases} Ax+By+c=0,\\ Ax+By+k=0
\end{cases} $$

If you subtract the second line from the firs, you obtain $c-k=0$. Then, if $c\ne k$, you have no solution: the lines are parallel.

share|improve this answer
    
(The second equation should be on a new line, I don't know why it is written on the same line as the first one) –  zar Sep 18 '10 at 19:45
1  
This is backwards. You have to prove that if Ax+By+c=0 and Dx+Ey+f = 0 are parallel then A=g*D and B=g*E, where g is not 0. –  Aryabhata Sep 18 '10 at 19:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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