The intersection point of L1:x=−3 and L2:x−y=0 is A(−3,−3).
The intersection point of L1:x=−3 and L3:3x+y=0 is B(−3,9).
The lines L2 and L3 intersect at the origin O(0,0).
The distances from the origin to A and B are:
OA=(−3)2+(−3)2=32
OB=(−3)2+92=310
The angle ∠AOB between the segments OA and OB can be determined using the dot product of vectors OA and OB:
OA⋅OB=(−3)(−3)+(−3)(9)=9−27=−18
Since the dot product is negative, cos(∠AOB)<0, which means ∠AOB is an obtuse angle. Therefore, the internal bisector of ∠AOB is the bisector of the obtuse angle between the lines L2 and L3.
According to the internal angle bisector theorem in △AOB, the bisector of ∠AOB divides the opposite side AB at point C in the ratio of the adjacent sides:
BCAC=OBOA=31032=51
Squaring both sides yields:
BC2AC2=51⇒AC2BC2=5
Thus, BC2:AC2=5:1.
Answer: 5:1