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Question

Straight line 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. 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+y9=0
C x7y+13=0
Given Lines
3x+4y=5----(1) and 4x3y=15------(2)
On comparing eq (1) and (2) with y=mx+c
m1=34,m2=43
m1m2=43×34=1
Hence given lines are perpendicular
So, line AB=AC SO a right angled isosceles Δ
Hence, the line BC through (1,2) will make angle of 450 with given lines
So, possible equation of BC are
(y2)=m±tan4501mtan450(x1)
where m= slope of AB34
(y2)=34±11(34)(x1)
(y2)=3±44±3(x1)
(y2)=17(x1) and (y2)=7(x1)
x7y+13=0 and 7x+y9=0

880239_190472_ans_8d0ab0045913447c80dbb90b1822d631.png

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