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Question

# Straight lines 3x+4y=5 and 4xâˆ’3y=15 intersect at the point A. Points B and C are chosen on these two lines such that AB=AC. Determine the possible equations of the line BC passing through the point (1,2).

A
x7y+13=0 and 7x+y9=0
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B
x+7y13=0 and 7xy9=0
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C
x7y+13=0 and 7xy+9=0
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D
x7y13=0 and 7x+y+9=0
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Solution

## The correct option is A x−7y+13=0 and 7x+y−9=0Let m1 and m2 be the slope of lines 3x+4y=5 and 4x−3y=15 respectivelyThen, m1=−34 and m2=43.Clearly, m1m2=−1So, lines AB and AC are at right angle. thus the triangle ABC is a right angles isosceles △Hence, the line BC through (1,2) will make an angle of 45O with the given lines.So, the possible equation of BC are(y−2)=m±tan45O1∓mtan45O(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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