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Question

# The equation of the straight line passing through the point of intersection of the lines x−y=1 and 2x−3y+1=0 and parallel to the line 3x+4y=14 is

A
3x+4y+24=0
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B
3x+4y24=0
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C
4x+3y+24=0
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D
4x+3y24=0
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Solution

## The correct option is B 3x+4y−24=0Equation of any line through the point of intersection of lines x−y−1=0 and 2x−3y+1=0 is (x−y−1)+λ(2x−3y+1)=0 ⇒(1+2λ)x−(1+3λ)y+(λ−1)=0⋯(1) Slope of line m1=1+2λ1+3λ Slope of line 3x+4y−14=0 is m2=−34 As both the lines are parallel, we get 1+2λ1+3λ=−34⇒4+8λ=−3−9λ⇒λ=−717 Putting the value of λ in equation (1), we get 317x+417y−2417=0∴3x+4y−24=0 Alternate solution: Line parallel to 3x+4y−14=0 is 3x+4y+μ=0 Finding the point of intersection of the other two lines, x−y−1=0 and 2x−3y+1=0 x=y+1⇒2(y+1)−3y+1=0⇒y=3⇒x=4 Now, 3×4+4×3+μ=0⇒μ=−24 Hence, the required equation of line is 3x+4y−24=0

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