A) \[7x+3y=0\]
B) \[7x-y=0\]
C) \[3x+2y=0\]
D) \[x+y=0\]
Correct Answer: A
Solution :
Let \[{{L}_{1}}=2x+y-1=0\] \[{{L}_{2}}=3x+2y-5=0\] We know that the equation of straight line passing through the intersection point of the lines \[{{L}_{1}}\] and \[{{L}_{2}}\] is given by \[{{L}_{1}}+\lambda {{L}_{2}}=0\] \[\Rightarrow \]\[(2x+y-1)+\lambda (3x+2y-5)=0\] Since, this line passes through the origin, also \[\therefore \] \[(0+0-1)+\lambda (0+0-5)=0\] \[\Rightarrow \] \[-\,5\lambda =1\] \[\Rightarrow \] \[\lambda =-\frac{1}{5}\] \[\therefore \]Required line is \[\left( 2x+y-1 \right)-1\left( 3x+2y-5 \right)=0\] \[\Rightarrow \]\[\left( 2-\frac{3}{5} \right)x+\left( 1-\frac{2}{5} \right)y-1+1=0\] \[\Rightarrow \] \[\frac{7}{5}x+\frac{3}{5}y=0\] \[\Rightarrow \] \[7x+3y=0\]
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