JEE Mains · Maths · STD 11 - 9. straight line
If the line, \(2 x-y+3=0\) is at a distance \(\frac{1}{\sqrt{5}}\) and \(\frac{2}{\sqrt{5}}\) from the lines \(4 x-2 y+\alpha=0\) and \(6 x-3 y+\beta=0,\) respectively, then the sum of all possible values of \(\alpha\) and \(\beta\) is
- A \(12\)
- B \(30\)
- C \(18\)
- D \(60\)
Answer & Solution
Correct Answer
(B) \(30\)
Step-by-step Solution
Detailed explanation
Apply distance between parallel line formula \(4 x-2 y+\alpha=0\) \(4 x-2 y+6=0\) \(\left|\frac{\alpha-6}{255}\right|=\frac{1}{55}\) \(|\alpha-6|=2 \Rightarrow \alpha=8,4\) \(\operatorname{sum}=12\) again \(6 x-3 y+\beta=0\) \(6 x-3 y+9=0\)…
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