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Question

# The equation of the internal bisector of ∠BAC of ΔABC with vertices A(5, 2), B(2, 3) and C(6, 5), is

A
2x + y + 12 = 0
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B
x + 2y – 12 = 0
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C
2x + y – 12 = 0
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D
x + 2y +12 = 0
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Solution

## The correct option is C 2x + y – 12 = 0Let AD be the internal bisector of angle BAC cutting BC at D. Now, AB=√(5−2)2+(2−3)2=√10and AC=√(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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