Shape induced by the condition of angle bisector

Hugh Soto

Hugh Soto

Answered question

2022-03-31

Shape induced by the condition of angle bisector passing through a fixed point
Let A, B be two points on 2D plane. For any CAB, define the set S={P;;APC=CPB}.
What is the shape of S?

Answer & Explanation

jmroberts70pbo2

jmroberts70pbo2

Beginner2022-04-01Added 10 answers

Step 1
Here is a solution using coordinate geometry. If |AB|=k and d|AC||BC|=t where CAB.
Using angle bisector theorem,
|AP||PB|=|AC||BC|=t|AP|2=t2|PB|2
WLOG, coordinates of A is (0,0) and of B is (k,0). If coordinates of P is (x,y),
we get x2+y2=t2((xk)2+y2)
Step 2
If t=1,x=k2 i.e. the locus is perpendicular bisector of AB.
If t1,~x2+y22kt2t21x=k2t2t21
(xkt2t21)2+y2=(ktt21)2
which is a circle with center at (kt2t21,0) and radius kt|t21|.

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