Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have the line $ax + bx + c = 0$. How do I find the values of $j$, $k$, and $l$ for the perpendicular line $jx + kx + l = 0$? Where the two lines cross at point $(p,q)$?

I know if the line were represented in slope-intercept form all I'd have to do is divide $-1$ by the slope and then multiply $p$ by the new slope and subtract the result from $q$ and make the difference the new value for $B$.

For example: The line perpendicular to $y = 4x + 3$, crossing at the point $(2,11)$ is

$m = -\frac{1}{4}$

$d = 2m = -\frac{1}{2}$

$11 - \left(-\frac{1}{2}\right) = 11.5$

therefore the equation for the perpendicular line is $y = -\frac{1}{4}x + 11.5$.

I'm not very familiar with general equations of lines and have been having a lot of trouble finding a good informational resource. How do I do the same thing for a line in the general equation form?

share|cite|improve this question
Did you mean $ax+by+c=0$? If so, you can rearrange it to $by = -ax - c$. If $b$ is nonzero then you can divide by it to obtain $y = -\frac{a}{b}x - \frac{c}{b}$, which is in slope-intercept form. – Brad Nov 6 '12 at 5:01
up vote 1 down vote accepted

Since the line $\,ax+by+c=0\,$ has slope equal to $\,-\dfrac{a}{b}\,\,,\,b\neq 0\,$ , any perpendicular line to it has t0 have slope equal to $\,\dfrac{b}{a}\,$ (why?) .

After the above, all you have to do is to choose any point on the given line.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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