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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?

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  • $\begingroup$ 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
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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 $d=(n_1+n_2)/2$.

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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.

enter image description here

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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