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A question reads, "Find equation of the line which is equidistant from parallel lines $9x + 6y – 7 = 0$ and $3x + 2y + 6 = 0$."

I solved it by doing the following:-

  1. Finding the distance between the two given lines, halved it to find the distance between the required line which is $9x + 6y + C = 0$ and the line $3x + 2y + 6 =0$.

    1. Took a point $(-2,0)$ lying on the line $3x + 2y + 6 = 0$.

    2. Used the formula for perpendicular distance between a point and a line, but due to modulus, I get two solutions for $C$ (and for the equation).

The two solutions for $C$ are $11/2$ (which is correct) and the other garbage value $61/2$ which is obviously wrong as it would lie above both the given equations as its $y$-intercept is larger than both the equations.

My question is - Why does this garbage value arrive?

Thank you.

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  • $\begingroup$ What exactly is the $C$ you're solving for here? You haven't defined it. $\endgroup$ – jmerry Feb 16 at 13:14
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    $\begingroup$ It seems like you have been looking for a line that is parallel to and a certain distance from one particular line. If that is the case, then of course you will get two lines: one above, and one below that particular line. $\endgroup$ – Minus One-Twelfth Feb 16 at 13:22
  • $\begingroup$ @jmerry sorry. I have corrected that now. Thank you. $\endgroup$ – Ram Keswani Feb 16 at 13:24
  • $\begingroup$ Beats me. When I try to set up a quadratic equation need on equal squared distances from the two lines, the leading coefficient vanishes and the "garbage" line is thus taken out to an infinitely distant curb. $\endgroup$ – Oscar Lanzi Feb 16 at 13:25
  • $\begingroup$ @MinusOne-Twelfth right, thank you $\endgroup$ – Ram Keswani Feb 16 at 13:27
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Visualise a set of railway tracks, its two metal lines being those whose equations you're given. In step 1, you worked out the track's half-width. In step 2, you marked a point on, say, the easternmost of the lines. In step 3, you moved perpendicular to that line through a half-width's worth. Going West takes you inside the track, halfway to the other line; going East juts out in the opposite direction.

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If you have two equations of parallel lines of the form

$$ax + by = c_1\tag{a}$$ and $$ax + by = c_2\tag{b}$$ then it is easy to see that the equidistant line between them has equation $$ax+by = \frac{c_1+c_2}{2}\tag{c}$$ (so just take the average of the right hand sides).

Rewrite your equations as

$$9x+6y=7 \tag{1}$$ $$9x+6y=-18\tag{2}$$

and apply the above. Then rewrite it to your favourite form.

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  • $\begingroup$ (+1) This was just what I was thinking of. $\endgroup$ – robjohn Feb 16 at 13:44
  • $\begingroup$ This is certainly a much simpler method than the OP’s (and kudos for pointing it out), but it doesn’t really answer the actual question being asked: why was there an extraneous solution in the OP’s approach? $\endgroup$ – amd Feb 16 at 19:52

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