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

Question

The equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ represents a pair of parallel lines. Prove that the equation of the line mid way between the two parallel lines us $hx+by+f=0$

My Attempt:

Let the lines be $lx+my+n_1=0$ and $lx+my+n_2=0$. Then, $$(lx+my+n_1)(lx+my+n_2)=0$$

Comparing the above equation with $ax^2+2hxy+by^2+2gx+2fy+c=0$, we get: $l^2=a, m^2=b, lm=h, l(n_1+n_2)=2g, m(n_1+n_2)$

Also,

distance between the two parallel lines represented by the given equation is $d=2 \sqrt {\frac {g^2 -ac}{a(a+b)}}$

$d=2 \sqrt {\frac {g^2 -ac}{h^2 +a^2}}$?

Now, what should I do to complete the proof?

Answer 1

Clarifing the last hint. The line mid is also parallel to the others lines, so it has the form: $$lx+my+d=0\tag{1}$$ but it lies in the middle, so $(0,-n_1/m)$ at the first line and $(0,-n_2/m)$ at the second line are both at the same vertical line and then the middle point $(0, -(n_1+n_2)/2m)$ must belongs to the line (1). Thats why we must have $d=(n_1+n_2)/2$.

Answer 2

The line mid way between parallel lines $lx+my+n_1=0$ and $lx+my+n_2=0$ is $$lx+my+\frac{n_1+n_2}{2}=0.$$

Why?

When considering parallel lines, the distance in any direction is enough to look for the mid line. You do not need to take the perpendicular one.

Thus, you can consider, for instance, the distance in the $y$-axis direction. Line $lx+my+n=0$ crosses the vertical axis at point $(0,−n/m)$. Of course, it is the same for $n_1$ and $n_2$.

So distances between any two of those points $(0,a)$ and $(0,b)$ is given by $|a-b|$. In particular, in the case of the parallel lines, the mid line needs to satisfy $|n_1-n|=|n-n_2|$.