If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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.

## Want to join the conversation?

• 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?
• 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.``
• When constructing parallel lines with a compass and straightedge, how should you start the construction?
• What software or app is this with this compass & straight edge?
• I am pretty sure it was embedded into the practice at the time.
(1 vote)
• 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?
• what is arbitrary point?
(1 vote)
• 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)
• Can you guess where the line will be, without useing circles?
(1 vote)
• Constructions are very precise, guessing may only give an approximate answer. You do not need full circles, just arcs where both sides intersect.
(1 vote)
• Where are the Construction Problems?
(1 vote)
• Unfortunately, there are none on KA. They should really make some though.
(1 vote)
• I Haley and I'm in the 10 I would have not thought of this math that they have this day
(1 vote)
• Oops, wrong tab, but great thing to hear!
(1 vote)
• How do i do this problem algebra bisects ? m<ABX=5x,m <XBC=3x+10
(1 vote)
• is a perpendicular line drawn from the given lie meant to be always the perpendicular bisector of the same line??
(1 vote)
• Perpendicular bisectors have to do with line segments because you cannot bisect a line. A perpendicular line or line segment can be drawn anywhere along a line segment, even at one of the end points.
(1 vote)

## 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.