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Geometric constructions: perpendicular line through a point on the line

Sal constructs a line perpendicular to a given line through a point on the line using compass and straightedge. Created by Sal Khan.

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  • hopper cool style avatar for user Christi
    In the video you used two circles that had the same radius. But you don't really need to have the radii equal to each other, do you?
    (15 votes)
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    • hopper cool style avatar for user Christi
      Actually, just looked at the hints for the problem (which does not use two congruent circles):

       How can you guarantee that a line is perpendicular?

      If we pick two points on the perpendicular line which are an equal distance from the intersection, they will also be the same distance from every other point on the line we started with.

      If we don't already have the perpendicular line, is there another way to find the blue points?

      If we use the compass to put a circle somewhere on the line, the circle will include all points that are the same distance from that point, including the two blue points.

      We can add a second circle somewhere else on the line that intersects with the first circle.
      (8 votes)
  • blobby green style avatar for user Jeffy Cherian
    When constructing parallel lines with a compass and straightedge, how should you start the construction?
    (3 votes)
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  • piceratops ultimate style avatar for user Mark Wilson
    What software or app is this with this compass & straight edge?
    (3 votes)
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  • orange juice squid orange style avatar for user PeterB
    When I try the exercises in "Compass constructions 1", my tools behave differently than Sal's. Specifically, my straightedge tool seems to only create line segments, not lines, which makes solving the problem I'm working on ("Copy an angle") extremely annoying. Does anyone else see this? Is there some user interface to change the straightedge mode that I'm overlooking?
    (2 votes)
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  • cacteye blue style avatar for user Chinmayee15
    what is arbitrary point?
    (1 vote)
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    • mr pink green style avatar for user David Severin
      Arbitrary just means random, so an arbitrary point is a point that is marked somewhere without much consideration. In this case, it would have to be on the paper and you would not want it too close to the line or the ends of the line, but that leaves a lot of space to put it in.
      (1 vote)
  • leaf red style avatar for user Philip Kujawa
    Can you guess where the line will be, without useing circles?
    (1 vote)
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  • mr pink red style avatar for user SFAIRFIELD
    Where are the Construction Problems?
    (1 vote)
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  • aqualine seed style avatar for user Haley Ellenburg
    I Haley and I'm in the 10 I would have not thought of this math that they have this day
    (1 vote)
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  • blobby green style avatar for user GarciaJoanne69
    How do i do this problem algebra bisects ? m<ABX=5x,m <XBC=3x+10
    (1 vote)
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  • piceratops ultimate style avatar for user Anwesha Mishra
    is a perpendicular line drawn from the given lie meant to be always the perpendicular bisector of the same line??
    (1 vote)
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Video transcript

Construct a line perpendicular to the given line. So if I can pick two arbitrary points on this line, and if I can make a line that is always equidistant from those two points, then that line will be perpendicular. And actually, it will be a perpendicular bisector of the segment formed by those two points. Now, they don't care whether we're bisecting anything. But they do care about it being perpendicular. So let's do this. So I'm going to draw a circle with my compass. And so let's just pick that point right over there. I could adjust the radius if I like. But I might as well-- well, I'll just leave it right over there. Now let me draw another circle. And this time, I'm going to center it where the first circle intersects with my line. And then I'm going to adjust the radius to overlap with the first dot. And now, where these two circles intersect, those are points that are equidistant from both of these centers that I just constructed. So let me draw a line that connects those two. And that line is going to be perpendicular to our original line.