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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 = 0
Let AD be the internal bisector of angle BAC cutting BC at D.

Now, AB=(52)2+(23)2=10and AC=(56)2+(25)2=10

since AD is the internal bisector of angle BAC,

BDDC=ABAC=1010=11

Coordinates of D are (2+62,3+52) i.e. (4, 4)

So, the equation of AD is

y2=2454 (x – 5) or 2x + y – 12 = 0


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