Question

To find the condition that the general equation of the second degree...

Answer 1

\[ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0\]

may represent two straight lines. And, hence, find the point of intersection.

Answer 2

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Answer 3

To find the condition for the second degree equation to represent pair of straight lines :

Let any general \(2^{nd}\) degree equation be :

\(ax^2 + 2hxy + 2gx + 2fy + c = 0 \)

Then :

|a h g|
|h b f| = 0
|g f c|

Or abc + 2fgh - \(af^2 -bg^2 -ch^2\) = 0

And \ (h^2 \ge ab\)

Answer 4

*\(h^2 \ge ab\)

Answer 5

Why the term by^2 is deleted from the main equation ?

Answer 6

http://books.google.com/books?id=tasLBU-GEzwC&pg=PA244&lpg=PA244&dq=how+can+ax2%2B2hxy%2Bby2%2B2gx%2B2fy%2Bc%3D0+be+2+straight+lines&source=bl&ots=KJKFkeeiwa&sig=x1o5um2j7Kk5U1tJeidJr7hTKFw&hl=en&sa=X&ei=2SfOUfeqLce3yQHE9YHICg&ved=0CE0Q6AEwBw#v=onepage&q=how%20can%20ax2%2B2hxy%2Bby2%2B2gx%2B2fy%2Bc%3D0%20be%202%20straight%20lines&f=false

Answer 7

That might help you.

Answer 8

Sorry forgot that.

Answer 9

\[\left| \begin{array}{ccc}
a & h & g\\
h & b &f\\
g & f & c\end{array} \right| =0 .\]

Answer 10

Why you wrote h^2 >= ab ?

Answer 11

Angle between two lines : \(\alpha \) :
\(\tan \alpha = \cfrac{2\sqrt{h^2-ab}}{a+b} \)

Implies : \(h^2 \ge ab\) If, \(h^2 < ab\) then \(\tan \alpha\) will not be defined.

Answer 12

@goformit100 you just pull the most random math question which you dont even try to work out with the user who is helping you. All you reply is "Ok" "Thank you Sir" but dont really work it out and collaborate to understand the material. Just to make it clear, i am not trying to offend you in any possible way, as i come in peace, but com'on!

Answer 13

I was offline