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Question

# A straight line passing through the point A(–2, –3) cuts the line x + 3y = 9 and x + y + 1 = 0 at B and C respectively. If AB.AC = 20, then equation of line can be

A

x – y = 1

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B

x – y + 1 = 0

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C

3x – y + 3 = 0

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D

3x – y = 3

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Solution

## The correct options are A x – y = 1 C 3x – y + 3 = 0 Equation of line through A is y+3=m(x+2) ....(1) Let point B is (x1, y1) ∴x1=−2+AB cosθ, y1=−3+AB sin θ Put (x1, y1) in x+3y=9 We get AB=20cos θ+3 sin θ similarly find AC=4cos θ+sin θ and use AB.AC=20 We get tanθ=1,3 put in (1)

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