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Question

# Straight line 3x+4y=5 and 4xā3y=15 intersect at the point A. Points B and C are chosen on these two lines such that AB=AC. Possible equation of BC passing through (1,2) can be

A
5x+2y9=0
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B
7x+y9=0
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C
x7y+13=0
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D
7x+3y13=0
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Solution

## The correct options are B 7x+y−9=0 C x−7y+13=0Given Lines3x+4y=5----(1) and 4x−3y=15------(2)On comparing eq (1) and (2) with y=mx+cm1=−34,m2=43m1m2=43×−34=−1Hence given lines are perpendicular So, line AB=AC SO a right angled isosceles ΔHence, the line BC through (1,2) will make angle of 450 with given lines So, possible equation of BC are (y−2)=m±tan4501∓mtan450(x−1)where m= slope of AB−−34⇒(y−2)=−34±11∓(−34)(x−1)⇒(y−2)=−3±44±3(x−1)⇒(y−2)=17(x−1) and ⇒(y−2)=−7(x−1)⇒x−7y+13=0 and ⇒7x+y−9=0

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