<s>[INST] Why does the angle between a Pair of Straight Lines depend only on the Homogeneous Part? The most general form of a quadratic is: $ax^2 + by^2 + 2gx + 2fy + c + 2hxy= 0 $ and that for a homogeneous second degree equation is : $ax^2 + by^2 + 2hxy=0$ 
I derived the formula $$\tan \theta = \left|\dfrac{2\sqrt{h^2 - ab}}{a+b}\right|$$ 
but later, author claimed that the same formula is applicable for the general form too. 
But I (and he too) derived it for the homogeneous case only. How can he claim this then?
 [/INST] For $2^\circ$ homogeneous equation $ax^2+2hxy+by^2=0$, we say $m_1+m_2=-\dfrac{2h}{b}$ where $m_1$ is the slope of first line and $m_2$ is the slope of second line.
Proof:
$$ax^2+2hxy+by^2=0$$
$$a+2h\dfrac{y}{x}+b\dfrac{y^2}{x^2}=0$$
$$bm^2+2hm+a=0$$
$$m_1+m_2=-\dfrac{2h}{b}$$
But if we talk about general $2^\circ$ equation $ax^2+2hxy+by^2+2gx+2fy+c=0$, then also we can say $m_1+m_2=-\dfrac{2h}{b}$
Here is the proof:
$ax^2+2hxy+by^2+2gx+2fy+c=0$ can be written as 
$$\left(x+\dfrac{hy-y\sqrt{h^2-ab}}{a}+c_1\right)\left(x+\dfrac{hy+y\sqrt{h^2-ab}}{a}+c_2\right)=0$$
Slope of first line is 
$$m_1=\dfrac{-a}{h-\sqrt{h^2-ab}}=-\dfrac{h+\sqrt{h^2+ab}}{b}$$
$$m_2=\dfrac{-a}{h+\sqrt{h^2-ab}}=-\dfrac{h-\sqrt{h^2+ab}}{b}$$
$$m_1+m_2=-\dfrac{2h}{b}$$
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