A) x + y + 3 = 0
B) x - y - 3 = 0
C) x-3y-5=0
D) x-3y+5=0
Correct Answer: D
Solution :
The equation of a straight line passing through the point of intersection of x - y +1 = 0 and 3x+y-5=0 is \[(x-y+1)+\lambda (3x+y-5)=0\] or \[x(3\lambda +1)+y(\lambda -1)-(5\lambda -1)=0\] ? (i) It is perpendicular to 3x + y - 5 = 0. \[\therefore \] \[-3\times \left( -\frac{3\lambda +1}{\lambda -1} \right)=-1\] \[\Rightarrow \] \[\lambda =\frac{-1}{5}\] On putting\[\lambda =-\frac{1}{5}\]in Eq. (i), we get x - 3y + 5 = 0 as the equation of the required line.
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