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Question

Straight lines 3x+4y=5 and 4x3y=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 x7y+13=0 and 7x+y9=0
Let m1 and m2 be the slope of lines 3x+4y=5 and 4x3y=15 respectively
Then, m1=34 and m2=43.
Clearly, m1m2=1
So, 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
(y2)=m±tan45O1mtan45O(x1)

where, m= slope of AB=34

(y2)=34±11(34)(x1)

(y2)=3±44±3(x1)

(y2)=17(x1) and (y2)=7(x1)

x7y+13=0 and 7x+y9=0

347772_258483_ans_1027df00e53e409c98fd747718651fc3.png

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