A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[1\]
D) \[\frac{4}{3}\]
Correct Answer: D
Solution :
Key Idea: Any line perpendicular to\[ax+by=c\]is\[bx-ay=k\]. Given line\[3x+y=3\] \[\therefore \]Line perpendicular to \[3x+y=3\]is\[x-3y=\lambda \] Also, it passes through\[(2,\,\,2)\] \[\Rightarrow \] \[2-6=\lambda \Rightarrow \lambda =-4\] \[\therefore \]Equation of line\[x-3y=-4\] ... (i) \[\therefore \]\[y\]intercept\[=\frac{-4}{-3}=\frac{4}{3}\] Alternate Method: Any line through\[(2,\,\,2)\] is \[(y-2)=m(x-2)\] ...(i) \[(\because \]lines are\[\bot )\] \[\therefore \] \[m=-\frac{1}{slope\,\,of\,\,line\,\,3x+y=3}\] \[\Rightarrow \] \[m=\frac{1}{-3}=\frac{1}{3}\] ? (ii) \[\therefore \]From (i) and (ii) Required line\[3(y-2)=(x-2)\] On putting \[x=0\], we get \[y=\frac{4}{3}\]is the \[y\] intercept.
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