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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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    @Deb: Yes. Try writing a,b,c etc in terms of l,m,n and use formulae you know.2011-04-21

1 Answers 1

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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.

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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](http://mathworld.wolfram.com/TriangleArea.html). 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.2011-04-24