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

Answer 1

Answer: (D)

14

Answer 2

By observing origin and P lies in same region.
L1​(0,0)>0;L1​(α,β)>0⇒3α−4β+12>0 1=∣∣​53α−4β+12​∣∣​
3α−4β+12=5......(1)
Similarly for L2​
L2​(0,0)>0;L2​(α,β)>0
1=∣∣​108α+6β+11​∣∣​⇒8α+6β+11=10.......(2)
Solving (1) and (2)
α=−2523​;β=100106​
100(α+β)=100(100−92​+100106​)=14