Let the line L1 : x+3=0 intersect the lines L2 : x-y=0 and L3 : 3x+y=0 at the points A and B, respectively. Let the bisector of the obtuse angle between the lines L2 and L3 intersect the line L1 at the point C. Then BC2 : AC2 is equal to:
- 5:1
- 1:5
- 2:3
- 3:2
Answer
(1) 5:1
Solution
L2 and L3 meet at O(0, 0).
x=-3 on L2: A(-3, -3); on L3: B(-3, 9)
OA OB=(-3)(-3)+(-3)(9)=-18<0 is obtuse.
So the bisector of the obtuse angle between L2 and L3 is the internal bisector of in ; it meets AB (on L1) at C.
By the angle bisector theorem: BCAC= OBOA= 9+81 9+9= 90 18= 5
BC2 : AC2=5 : 1
(Check: the bisector 5(x-y)=3x+y meets x=-3 at C(-3, 3 5-6), giving AC=3( 5-1), BC=3 5( 5-1).)
Answer: (1)