The correct option is B 3x+4y−24=0
Equation 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