# Homogeneous second degree equation problem

We are given a curve having the equation:

$ax^2+2hxy+by^2+2gx+2fy+c=0$

and a line having the equation:

$lx+my=n$

$\frac{lx+my}{n} = 1$

While making the equation of the curve "homogeneous", we multiply all the terms having degree one with 1 and the terms having degree 0 with $1^2$. Then we substitute the value of 1 as shown in the above equation. And get the result

$ax^2+2hxy+by^2+2gx(\frac{lx+my}{n})+2fy(\frac{lx+my}{n})+c(\frac{lx+my}{n})^2 = 0$

But why do we do this? Is there any concrete reason or theorem supporting this?

• Could you provide some clarification? Are you asking why this is a valid operation? Are you asking why this is an interesting step in your problem of interest? – Michael Burr Oct 2 '18 at 17:59
• @MichaelBurr I am asking why this is a valid operation. – S.Nep Oct 2 '18 at 18:07

Therefore, at an intersection point, the $$(x,y)$$ will satisfy $$\frac{lx+my}{n}=1$$ as well as the formula for the conic. Therefore, at an intersection point, you can make the substitution $$\frac{lx+my}{n}=1$$ to construct a homogeneous equation in $$x$$ and $$y$$.