Question:
Lines are drawn parallel to the line \(4x-3y+2=0\), at a distance \(\dfrac35\) from the origin. Then which one of the following points lies on any of these lines ?
Lines are drawn parallel to the line \(4x-3y+2=0\), at a distance \(\dfrac35\) from the origin. Then which one of the following points lies on any of these lines ?
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Parallel lines are \(4x-3y+k=0\) with \(|k|/5=3/5\), so \(4x-3y=\pm3\).
Updated On: Nov 5, 2026
- \(\left(\frac14,-\frac13\right)\)
- \(\left(-\frac14,\frac23\right)\)
- \(\left(-\frac14,-\frac23\right)\)
- \(\left(\frac14,\frac13\right)\)
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The Correct Option is B
Solution and Explanation
A line parallel to the given one is \(4x-3y+k=0\). Its distance from the origin is \(\frac{|k|}{\sqrt{16+9}}=\frac{|k|}{5}=\frac35\Rightarrow k=\pm3\).
So the lines satisfy \(4x-3y=\pm3\). Evaluate \(4x-3y\) at the options:
(1) \(1+1=2\); (2) \(-1-2=-3\); (3) \(-1+2=1\); (4) \(1-1=0\).
Only option (2) gives \(\pm3\).
Answer: (2) \(\left(-\frac14,\frac23\right)\)
So the lines satisfy \(4x-3y=\pm3\). Evaluate \(4x-3y\) at the options:
(1) \(1+1=2\); (2) \(-1-2=-3\); (3) \(-1+2=1\); (4) \(1-1=0\).
Only option (2) gives \(\pm3\).
Answer: (2) \(\left(-\frac14,\frac23\right)\)
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