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CCSS.Math:

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.