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Question

The equation of the straight line passing through the point of intersection of the lines xy=1 and 2x3y+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+4y24=0
Equation of any line through the point of intersection of lines
xy1=0 and 2x3y+1=0 is
(xy1)+λ(2x3y+1)=0
(1+2λ)x(1+3λ)y+(λ1)=0(1)
Slope of line
m1=1+2λ1+3λ
Slope of line 3x+4y14=0 is
m2=34
As both the lines are parallel, we get
1+2λ1+3λ=344+8λ=39λλ=717

Putting the value of λ in equation (1), we get
317x+417y2417=03x+4y24=0


Alternate solution:
Line parallel to 3x+4y14=0 is
3x+4y+μ=0
Finding the point of intersection of the other two lines,
xy1=0 and 2x3y+1=0
x=y+12(y+1)3y+1=0y=3x=4
Now,
3×4+4×3+μ=0μ=24
Hence, the required equation of line is
3x+4y24=0

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