The vertices of a triangle are A(−1,−7),B(5,1),and C(1,4). The equation of the bisector of ∠ABC is ___
A
x-7y+2=0
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B
7x+y-2=0
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C
2y-7x-5=0
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D
2x+7y+5=0
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Solution
The correct option is A x-7y+2=0 Let BD be the bisector of ∠ABC then,
AD:DC=AB:BCandAB=√(5+1)2+(1+7)2=10BC=√(5−1)2+(1−4)2=5∴AD:DC=2:1 Therefore, by section formula, D=(13,13) therefore the equation of BD is y−1=13−113−5(x−5)ory−1=−23−143(x−5)or7y−7=x−5x−7y+2=0