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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 :

$-$14
42
$-$22
14

Solution

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