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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)

best_answer

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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