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Question

# Straight lines x–y=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 = 0Let the slope of line AC be m Then equation of AC be y−(−7)x−2=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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