A(1,3) and C(−2/5,−2/5) are the vertices of a triangle ABC and the equation of the internal angle bisector of ∠ABC is x+y=2, then
A
BC:7x−3y+4=0
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B
BC:7x+3y+4=0
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C
B≡(5/2,9/2)
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D
B≡(−5/2,9/2)
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Solution
The correct options are BBC:7x+3y+4=0 DB≡(−5/2,9/2) Let the image of A(1,3) in line x+y=2 be D(a,b). Here the midpoint of AD which is (a+12,b+32) lies on the line x+y=2.
Substituting this point in the line equation x+y=2, we get (a,b)=(−1,1)
So line BC passes through (−1,1) and (−2/5,−2/5).
The equation of line BC is y−1=−2/5−1−2/5+1(x+1) ⇒7x+3y+4=0
Vertex B is point of intersection of 7x+3y+4=0 and x+y=2, i.e., B≡(−5/2,9/2)