Lines are drawn parallel  to the line 4x-3y+2=0, at a 

distance \frac{3}{5} from the origin. Then which one of the following points

lies on any of these lines?

  • Option 1)

    (\frac{-1}{4},\frac{2}{3})

  • Option 2)

    (\frac{1}{4},\frac{-1}{3})

  • Option 3)

    (\frac{1}{4},\frac{1}{3})

  • Option 4)

    (\frac{-1}{4},\frac{-2}{3})

 

Answers (1)

Line parallel to 4x-3y+2=0 can be written as, 4x-3y+c=0.

This line lies at a distance of \frac{3}{5}  from origin.

=>\left |\frac{0-0-c}{\sqrt{16+9}} \right |=\frac{3}{5}

=>c=\pm 3

So, equation of lines are 4x-3y+3=0 and 4x-3y-3=0

Now, putting all options in these equations

(1)(\frac{-1}{4},\frac{2}{3})    

4(\frac{-1}{4})-3\cdot \frac{2}{3}+3=-1-2+3=0

4(\frac{-1}{4})-3\cdot \frac{2}{3}-3=-6 \neq 0

    

(2) (\frac{1}{4},\frac{-1}{3})

4(\frac{1}{4})-3\cdot \frac{-1}{3}+3=5\neq0

4(\frac{1}{4})-3\cdot \frac{-1}{3}-3=-1\neq0

 

(3) (\frac{1}{4},\frac{1}{3})

4(\frac{1}{4})-3\cdot \frac{1}{3}+3\neq0

4(\frac{1}{4})-3\cdot \frac{1}{3}-3\neq0

 

(4) (\frac{-1}{4},\frac{-2}{3})

4(\frac{-1}{4})-3\cdot \frac{-2}{3}+3\neq0

4(\frac{-1}{4})-3\cdot \frac{-2}{3}-3\neq0


Option 1)

(\frac{-1}{4},\frac{2}{3})

Option 2)

(\frac{1}{4},\frac{-1}{3})

Option 3)

(\frac{1}{4},\frac{1}{3})

Option 4)

(\frac{-1}{4},\frac{-2}{3})

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