If A(1,3) and C(−25,−25) are the vertices of a △ABC and the equation of the internal angle bisector of ∠ABC is x+y=2, then
A
the equation of line BC is 7x−3y+4=0
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B
the equation of line BC is 7x+3y+4=0
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C
B≡(52,92)
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D
B≡(−52,92)
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Solution
The correct options are B the equation of line BC is 7x+3y+4=0 DB≡(−52,92) Let the image of A(1,3) w.r.t. x+y=2 be D(a,b), so a−11=b−31=−2(1+3−2)2⇒D=(−1,1)
So equation of line BC is y−1=−25−1−25+1(x+1)⇒7x+3y+4=0
Therefore, vertex B is point of the intersection of 7x+3y+4=0 and x+y=2 i.e. B≡(−52,92)