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

Answer 1

By observing origin and P lies in same region.

Solving (1) and (2)

\(\alpha=-\frac{23}{25};\beta=\frac{106}{100}\)

\(100(\alpha+\beta)=100(\frac{-92}{100}+\frac{106}{100})=14\)