JEE Mains · Maths · STD 11 - 9. straight line
Let the point \(P(\alpha, \beta)\) be at a unit distance from each of the two lines \(L_{1}: 3 x-4 y+12=0\), and \(L _{2}: 8 x+6 y+11=0\). If \(P\) lies below \(L _{1}\) and above \(L_{2}\), then \(100(\alpha+\beta)\) is equal to
- A \(-14\)
- B \(42\)
- C \(-22\)
- D \(14\)
Answer & Solution
Correct Answer
(D) \(14\)
Step-by-step Solution
Detailed explanation
\(L_{1}: 3 x-4 y+12=0\) \(L_{2}: 8 x+6 y+11=0\) Equation of angle bisector of \(L_{1}\) and \(L_{2}\) of angle containing origin \(2(3 x-4 y+12)=8 x+6 y+11\) \(2 x+14 y-13=04\) \(\frac{3 \alpha-4 \beta+12}{5}=14\) \(3 \alpha-4 \beta+7=0\) Solution of \(2 x+14 y-13=0\) and…
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