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)

By observing origin and $P$ lies in same region.
$L _1(0,0)>0 ; L _1(\alpha, \beta)>0 \Rightarrow 3 \alpha-4 \beta+12>0$ $1=\left|\frac{3 \alpha-4 \beta+12}{5}\right|$
$3 \alpha-4 \beta+12=5......$(1)
Similarly for $L _2$
$L _2(0,0)>0 ; L _2(\alpha, \beta)>0$
$1=\left|\frac{8 \alpha+6 \beta+11}{10}\right| \Rightarrow 8 \alpha+6 \beta+11=10 .......$(2)
Solving (1) and (2)
$\alpha=-\frac{23}{25} ; \beta=\frac{106}{100}$
$100(\alpha+\beta)=100\left(\frac{-92}{100}+\frac{106}{100}\right)=14$