JEE Mains · Maths · STD 11 - 9. straight line
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?
- A \(\left( { - \frac{1}{4},\frac{2}{3}} \right)\)
- B \(\left( { \frac{1}{4},\frac{1}{3}} \right)\)
- C \(\left( { \frac{1}{4},-\frac{1}{3}} \right)\)
- D \(\left( { - \frac{1}{4}, - \frac{2}{3}} \right)\)
Answer & Solution
Correct Answer
(A) \(\left( { - \frac{1}{4},\frac{2}{3}} \right)\)
Step-by-step Solution
Detailed explanation
Required line is \(4x - 3y + \lambda = 0\) \(\left| {\frac{\lambda }{5}} \right| = \frac{3}{5}\) \( \Rightarrow \lambda \pm 3\) So, required equation of the line is \(4x - 3y + 3 = 0\) and \(4x - 3y - 3 = 0\)…
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