A) \[2x-3y+1=0\]
B) \[2x-3y+3=0\]
C) \[2x-3y+5=0\]
D) \[2x-3y+7=0\]
Correct Answer: D
Solution :
The point of intersection of lines\[x+3y-1=0\] and\[x-2y+4=0\]is ( -2 ,1). Let equation of line perpendicular to the given line is\[2x-3y+\lambda =0\]. Since, it passes through (-2 , 1). \[\therefore \] \[2(-2)-3(1)+\lambda =0\] \[\Rightarrow \] \[\lambda =7\] \[\therefore \]Required line is\[2x-3y+7=0\]You need to login to perform this action.
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