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.

  • $\begingroup$ What exactly is the $C$ you're solving for here? You haven't defined it. $\endgroup$ – jmerry Feb 16 at 13:14
  • 1
    $\begingroup$ 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

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.


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.

  • $\begingroup$ (+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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.