JEE Math Practice Question
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
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