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: $l^2=a, m^2=b, lm=h, l(n_1+n_2)=2g, m(n_1+n_2)$


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?

  • $\begingroup$ Your final equation should be $lx+my+\frac{n_1+n_2}{2} = 0$. Multiply this equation by $m$ and substitute the values as per the equations that you derived. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 23 '16 at 14:08
  • $\begingroup$ @expiTT, how do I get there? And what is the equation derived? $\endgroup$ – pi-π Nov 23 '16 at 14:13
  • $\begingroup$ Use distance formula between two parallel lines and check that the parallel line (say $lx+my+n=0$) midway between the two parallel lines in your explanation must satisfy $|n_1-n|=|n-n_2|$. This will give you $n=\frac{n_1+n_2}{2}$. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Nov 23 '16 at 14:24
  • $\begingroup$ Hint: you’ll need to multiply the equation $lx+my+(n_1+n_2)/2=0$ by some constant factor to get it into the desired form. $\endgroup$ – amd Nov 23 '16 at 17:45

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 $d=(n_1+n_2)/2$.


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


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.

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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 $|n_1-n|=|n-n_2|$.


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