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If $$ax^2+2hxy+by^2+2gx+2fy+c=0,$$ represents a pair of lines, show that the square of the distance from origin to their point of intersection is $$\frac{c(a+b)-f^2-g^2}{ab-h^2}.$$

My attempts; since the given equation represents a pair of straight lines, let each be $$l_1x+m_1y+n_1=0$$ and $$l_2x+m_2y+n_2=0.$$ Now comparing the combined equation of these two lines with the given equation I got: $$l_1l_2=a$$ $$m_1m_2=b$$ $$n_1n_2=c$$ $$l_1m_2 + l_2m_1=2h$$ $$l_1n_2 +l_2n_1=2g$$ $$m_1n_2+ m_2n_1=2f.$$ Now how should I complete?

Iaamuser user
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  • While this is likely to get a little bogged down in calculation (much like the other approaches), I would consider first rotating to a coordinate system that's missing the cross term (since this won't change the distance but may make some of the subsequent calculations easier). – Steven Stadnicki Feb 08 '16 at 17:50

2 Answers2

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Step $1:$ Calculate the solution $(x_0,y_0)$ for the system $$ l_jx+m_jy+n_j=0,\qquad j=1,2 $$ Using Crammer's rule we get $$ x_0^2=\left({ m_1n_2-m_2n_1\over l_1m_2-l_2m_1}\right)^2={( m_1n_2+m_2n_1)^2-4n_1n_2m_1m_2\over(l_1m_2+l_2m_1)^2-4l_1l_2m_1m_2}={f^2-bc\over h^2-ab } $$ Similarly calculate $$y_0^2={g^2-ac\over h^2-ab}$$.

Step $2:$ Calculate $x_0^2+y_0^2$.

Jack's wasted life
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Hint they represent a pair when $abc+2fgh=af^2+bg^2+ch^2$ can you continue from here

Archis Welankar
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