aroub
  • aroub
I really need some help in how to find or know the locus >.< Given a fixed angle CAB. Find the locus of the points equidistant from the sides of
Mathematics
schrodinger
  • schrodinger
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phi
  • phi
equidistant from two sides of an angle is the angle bisector 5 cm from a point is the circumference of the circle with center at point B with radius 5 the intersection of the circle and the angle bisector is the answer: 2 points
phi
  • phi
though if the circle is too small, no intersection, or possibly 1 tangent point....
aroub
  • aroub
How on earth do you find these things?! It's sooo confusing

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phi
  • phi
You memorize (or learn) the various cases. we can prove the angle bisector gives you the locus of points equidistant for both sides (we use congruent right triangles) It is (more or less) intuitive that the locus of points equidistant from a point is a circle (this is the definition of a circle)
aroub
  • aroub
And I don't think they want to know how many points.. =) I've memorized the cases or conditions. It's sometimes it's hard to find it =( But just one question on this, how did we know that point B is the center?
aroub
  • aroub
( mind the second "it's" :P )
phi
  • phi
distance of 5 cm from point B the locus of all points 5 cm from B is a circle The actual answer to this question could be |dw:1352048840615:dw| depends on how "big" 5 cm is. we need more info to know if the circle intersects the bisector line
aroub
  • aroub
Well, it's not construction so it doesn't matter right now. But I will go with the first one.
aroub
  • aroub
I'm posting another few questions.. If you would please help me =) Because you and Hero are like the most people that I know are VERY good in geometry.

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