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This is rather a continuation for this,but this is much precise.After proving and understanding the basic formulas for pair of straight lines I am having some troubles with these:

  • If the equation $ax^2+by^2+2hxy+2gx+2fy+c=0$ represents a pair of parallel lines if $h^2 = ab$ and $bg^2=af^2$,then the distance between the parallel lines is $\large 2\sqrt{\frac{g^2-ac}{a^2+ab}}$ or $\large 2\sqrt{\frac{f^2-ac}{b^2+ab}}$.

  • The area of the triangle formed $ax^2+2hxy+by^2=0$ and $lx+my+n=0$ is $ \large \frac{n^2\sqrt{h^2-ab}}{|am^2-2hlm+bl^2|}$

In my module no proof is given just given as formula,I am very much interested to know how could we prove them?

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Quixotic
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  • What have you tried? Did you try picking up a book on co-ordinate geometry? The book by SL Loney might have it. This answer is most likely tedious algebra... – Aryabhata Apr 21 '11 at 20:26
  • @Moron:I haven't checked SL lonely but I did checked my book unfortunately it was not inculed in our boards syllabus so nothing there :/ – Quixotic Apr 21 '11 at 20:34
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    @Deb: The first one would be just setting your equation to $(lx + my +n)(lx + my + r)$. Did you try that? – Aryabhata Apr 21 '11 at 20:36
  • @Moron:Yes,then the distance would be $\large |\frac{n-r}{\sqrt{l^2+m^2}}|$ but what I am not getting how could we reduce this into that? – Quixotic Apr 21 '11 at 20:56
  • @Deb: Did you try equating the coefficients? – Aryabhata Apr 21 '11 at 20:58
  • Yes,I just now did that but yet I couldn't understand the crux of why I am doing that or may be I am missing something,I think you mean to compute the product and find value of a,b,c,g,and f ? – Quixotic Apr 21 '11 at 21:08
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    @Deb: Yes. Try writing a,b,c etc in terms of l,m,n and use formulae you know. – Aryabhata Apr 21 '11 at 21:09

1 Answers1

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1) Multiply by $a$ (for a nicer computation) and write

$a^2x^2+aby^2+2haxy+2gax+2fay+ac= (lx+my+n)(lx+my+r)$

You get $l=a$, $m=\pm h$, $r+n=2g$, $r+n=\pm 2fa/h$, $nr=ac$.

To proceed you need $2g=\pm 2fa/h$ which is equivalent to $g^2=f^2 a^2/(ab)$ which is given. So $r,n = g \pm \sqrt{g^2- ac}$.

Now use your formula for the distance of parallel lines.

2) Notice that $ax^2+2hxy+by^2=0$ is equivalent to $a^2x^2+2ahxy+h^2y^2=h^2y^2-aby^2=0$ so get three lines $lx+my+n=0$, $ax+hy=\sqrt{h^2-ab}y$ and $ax+hy=-\sqrt{h^2-ab}y$. You probably have a formula for calculating the area of this triangle.

J. M. ain't a mathematician
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Phira
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  • I wrote "2." for the paragraph starting with Notice, but the software writes "1." while in the edit window I still have "2.". Does anyone know why this is? – Phira Apr 23 '11 at 10:35
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    Numbering markdown acts weirdly if the numbers are separated by a lot of text. Anyway... @Deb: to add to the hint for #2, you should be able to find out the vertices of the triangle formed by the lines, and then use formula 16 in here. For the first, you could try rotating coordinates to remove the cross term $2hxy$, and then have your "parallel lines" be parallel to any of the two axes. From there, it should be easy to figure out the distance of two parallel lines that are parallel to a coordinate axis. – J. M. ain't a mathematician Apr 24 '11 at 16:36