# How do I find a perpendicular line from an equation in general form?

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?

• 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. – bradhd Nov 6 '12 at 5:01

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?) .