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Question

Three vertices of triangle ABC are A(1,11), B(9,8) and C(15,2). The equation of angle bisector of angle A is

A
(82771473)y+(3277+973)x91277+16373=0
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B
(82771473)y(3277973)x91277+16373=0
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C
(82771473)y+(3277973)x91277+16373=0
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D
(8277+1473)y+(3277973)x91277+16373=0
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Solution

The correct option is C (82771473)y+(3277973)x91277+16373=0
Given Points
A(1,11),B(9,8),C(15,2)
Equation of side AB
y11=81191(x1)
y11=38(x1)
8(y11)=3(x1)
8y88=3x+3
3x+8y91=0---(1)
Equation of side AC
y11=211151(x1)
y11=914(x1)
14(y11)=9(x1)
14y154=9x+9
9x+14y163=0---(2)
Eq of angle bisector of AB,AC
3x+8y9132+82=±(9x+14y16392+142)
3x+8y919+64=±(9x+14y16381+196)
3x+8y9173=±(9x+14y163277)
On taking positive sign
3x+8y9173=(9x+14y163277)
277(3x+8y91)=73(9x+14y163)
3277x+8277y91277=973x+1473y16373
(82771473)y+(3277973)x91277+16373=0

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