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Question

Straight lines xy=7 and x+4y=2 intersect at B. Points A and C are so chosen on these two lines such that AB = AC. The equation of line AC passing through (2, –7) is

A
x – y – 9 = 0
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B
23x + 7y + 3 = 0
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C
2x – y – 11 = 0
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D
7x – 6y – 56 = 0
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Solution

The correct option is B 23x + 7y + 3 = 0
Let the slope of line AC be m
Then equation of AC be y(7)x2=m
Let the angle between line AB and BC be θ
tanθ=∣ ∣1(14)1+1(14)∣ ∣
tanθ=∣ ∣5434∣ ∣
tanθ=53
Since AB=AC if we consider ABC
ABC=ACB
Equating tangent of both angles we get
53=∣ ∣m(14)1+m(14)∣ ∣
53=m(14)1+m(14) or 53=m(14)1+m(14)
Solving we get
m=1,237
m won't be 1 because we will get our AB line in that case.
So m=237

Equation of line is 23x + 7y + 3 = 0

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