    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
xy9=0
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B
23x+7y+3=0
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C
2xy11=0
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D
7x6y56=0
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Solution

## The correct option is B 23x+7y+3=0 Let the slope of the line AC be m. Then, the equation of AC be y+7=m(x−2). Slope of the line x−y=7 is 1 and slope of the line x+4y=2 is −14. Let the angle between line AB and BC be θ. ∴tanθ=∣∣ ∣ ∣ ∣∣1−(−14)1+1(−14)∣∣ ∣ ∣ ∣∣ ⇒tanθ=53 In △ABC, AB=AC ⇒∠ABC=∠ACB=θ ⇒tan(∠ABC)=tan(∠ACB) ⇒53=∣∣ ∣ ∣ ∣∣m−(−14)1+m(−14)∣∣ ∣ ∣ ∣∣ ⇒53=±⎛⎜ ⎜ ⎜⎝m+141−m4⎞⎟ ⎟ ⎟⎠ On solving, we get m=1,−237 m≠1 because if m=1, then we get the line which is parallel to x−y=7 So, m=−237 ∴ Equation of line AC is 23x+7y+3=0  Suggest Corrections  1      Similar questions
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