Question
The equation of the internal bisector of ∠BAC of ΔABC with vertices A(5, 2), B(2, 3) and C(6, 5), is
Answer: Option C
:
C
Let AD be the internal bisector of angle BAC cutting BC at D.
Now,AB=√(5−2)2+(2−3)2=√10andAC=√(5−6)2+(2−5)2=√10
since AD is the internal bisector of angle BAC,
∴BDDC=ABAC=√10√10=11
∴ Coordinates of D are (2+62,3+52) i.e. (4, 4)
So, the equation of AD is
y−2=2−45−4 (x – 5) or 2x + y – 12 = 0
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:
C
Let AD be the internal bisector of angle BAC cutting BC at D.
Now,AB=√(5−2)2+(2−3)2=√10andAC=√(5−6)2+(2−5)2=√10
since AD is the internal bisector of angle BAC,
∴BDDC=ABAC=√10√10=11
∴ Coordinates of D are (2+62,3+52) i.e. (4, 4)
So, the equation of AD is
y−2=2−45−4 (x – 5) or 2x + y – 12 = 0
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