I have several $2d$ line segments. for example, if I take a one line segment having end points $(x_1, y_1)$ and $(x_2, y_2)$. Then, I want to make a perpendicular line which passes through the midpoint of that line segment. for the simplicity if I say mid point $(x_3, y_3)$ then, how could I derive.

Any idea please. Thanks.


1 Answer 1


The midpoint of the line segment through $(x_1,y_1)$ and $(x_2,y_2)$ is $\left({x_1+x_2\over 2},{y_1+y_2\over 2}\right)$.

Also, perpendicular lines have negative reciprocal slopes, so if you want a line perpendicular to a line with slope $m$, you want to use a slope of $-{1\over m}$.

Finally, use the point-slope formula for the equation of a line through, say, $(x_3,y_3)$ with slope $m$: $y-y_3=m(x-x_3)$.

So, for example, if you want the perpendicular bisector to the segment through $P(x_1,y_1)$ and $Q(x_2,y_2)$:

  1. The midpoint is $\left({x_1+x_2\over 2},{y_1+y_2\over 2}\right)$.
  2. The slope of $\overline{PQ}$ is $m={y_2-y_1\over x_2-x_1}$, so the slope of a perpendicular is $-{1\over m}={x_1-x_2\over y_2-y_1}$.
  3. Finally, the equation of the perpendicular bisector of $\overline{PQ}$ is then $$y- {y_1+y_2\over 2}={x_1-x_2\over y_2-y_1}\left(x-{x_1+x_2\over 2}\right).$$
  • $\begingroup$ as this line should passes through mid point, then should I substitute {(x1+x2)/2,(y1+y2)/2} instead of (x1, y1)? Also, need a another clarification. as my constructor of the line segment is in the form of r = x.cos(theta)+y.sin(theta), then what would be the my perpendicular line passing through the mid point. plz teach me. thanks $\endgroup$
    – gnp
    Feb 17, 2013 at 20:37
  • $\begingroup$ thanks for the edit. if you could please tell me how this can be done with polar coordinates as my constructor is in that way. $\endgroup$
    – gnp
    Feb 17, 2013 at 20:39
  • $\begingroup$ The key to converting to polar coordinates $(r,\theta)$ is that the polar form of a line with slope $m$ and $y$-intercept $b$ is $r\sin\theta=mr\cos\theta+b$. With a little algebra you can translate what I gave above to the polar situation. $\endgroup$
    – JohnD
    Feb 17, 2013 at 20:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .