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### Course: Integrated math 1 > Unit 17

Lesson 8: Constructing lines & angles- Geometric constructions: congruent angles
- Geometric constructions: parallel line
- Geometric constructions: perpendicular bisector
- Geometric constructions: perpendicular line through a point on the line
- Geometric constructions: perpendicular line through a point not on the line
- Geometric constructions: angle bisector
- Justify constructions

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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.

## 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?(16 votes)
- 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.(9 votes)

- What software or app is this with this compass & straight edge?(5 votes)
- I am pretty sure it was embedded into the practice at the time.(6 votes)

- When constructing parallel lines with a compass and straightedge, how should you start the construction?(3 votes)
- Can you guess where the line will be, without useing circles?(3 votes)
- Constructions are very precise, guessing may only give an approximate answer. You do not need full circles, just arcs where both sides intersect.(3 votes)

- I don't see any exercise for this. Was it discontinued or something?(1 vote)
- My guess is that this was old old old Khan.(6 votes)

- what is arbitrary point?(2 votes)
- 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.(3 votes)

- What is the source of these questions? From which website?(2 votes)
- Towards the end of the video, Sal says that the line intersecting the original line, will be perpendicular to our original line, what does he mean by this?(2 votes)
- Two lines are perpendicular when they intersect at a right angle.(3 votes)

- Where are the Construction Problems?(2 votes)
- Unfortunately, there are none on KA. They should really make some though.(2 votes)

- construct a triangle ABC in which bc=4cm, angleb=degaries, and ab-bc =2cm(2 votes)
- Please clarify what you mean by angleb= degaries. I assume that you meant to use some degree measure, but you did not put a number into it - 30°, 5°, or what.(2 votes)

## 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.