Lines are drawn parallel to the line 4x – 3y + 2 = 0, at a distance $${3 \over 5}$$ from the origin. Then which one of the following points lies on any of these lines ?
JEE · Math · previous-year question
- A.$$\left( {{1 \over 4}, - {1 \over 3}} \right)$$
- B.$$\left( { - {1 \over 4},{2 \over 3}} \right)$$correct
- C.$$\left( { - {1 \over 4}, - {2 \over 3}} \right)$$
- D.$$\left( {{1 \over 4},{1 \over 3}} \right)$$
Answer
B. $$\left( { - {1 \over 4},{2 \over 3}} \right)$$
Explanation
Line parallel to 4x – 3y + 2 = 0 is given as 4x – 3y + $$\lambda $$ = 0 distance from origin is $$\left| {{\lambda \over 5}} \right| = {3 \over 5}$$ $$ \Rightarrow \lambda = \pm 3$$ $$ \therefore $$ required lines are 4x – 3y + 3 = 0 & 4x – 3y – 3 = 0 By Putting $$\left( { - {1 \over 4},{2 \over 3}} \right)$$ on both lines it satisfy.
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