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Let the point $\mathrm{P(\alpha, \beta)}$ be at a unit distance from each of the two lines $\mathrm{L_{1}: 3 x-4 y+12=0}$ , and $\mathrm{L_{2}: 8 x+6 y+11=0}$ . If $\mathrm{P}$ lies below $\mathrm{L_{1}}$ and above $\mathrm{L_{2}}$ , then $\mathrm{100(\alpha+\beta)}$ is equal to
Option: 1
$-14$

Option: 2
$42$

Option: 3
$-22$

Option: 4
$14$

Answers (1)

best_answer

P and origin lie on same side of $\mathrm{L_{1}}$

$\mathrm{\Rightarrow \frac{3 \alpha-4 \beta+12}{12}>0} \\$

$\mathrm{\Rightarrow 3 \alpha-4 \beta+12>0}$ ........(i)

Similarly for $\mathrm{L_{2}}$

$\mathrm{\frac{8 \alpha+6 \beta+11}{11}>0} \\$

$\mathrm{\Rightarrow 8 \alpha+6 \beta+11>0}$ ..........(ii)

Also distance from $\mathrm{L_{1}}$ and $\mathrm{L_{2}=1}$

$\mathrm{\Rightarrow \frac{|3 \alpha-4 \beta+12|}{5}=1} \\$

$\mathrm{\Rightarrow 3 \alpha-4 \beta+12=5}$ ( using (ii) )

And $\mathrm{\frac{|8 \alpha+6 \beta+11|}{10}=1 }\\$

$\mathrm{8 \alpha+6 \beta+11=10}$

Solving these 2 equations

$\mathrm{\alpha=\frac{-23}{25}, \quad \beta=\frac{53}{50}} \\$

$\mathrm{\therefore 100\left(\alpha_{1} \beta\right)=14}$

Hence correct option is 4

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