Let $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right), \left({\mathrm{x}}_2,{\mathrm{y}}_2\right), \left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ and $\left({\mathrm{x}}_4,{\mathrm{y}}_4\right)$ are four points which are at unit distance from the lines $3\mathrm{x}-4\mathrm{y}+1=0$ and $8\mathrm{x}+6\mathrm{y}+1=0,$ then the value of $\frac{\sum _{\mathrm{i}=1}^4{\mathrm{x}}_i}{\sum _{\mathrm{i}=1}^4{\mathrm{y}}_i}$ is equal to

• A

  $2$
• Correct answer
  B

  $–2$
• C

  $1$
• D

  $–1$

Solution

Point of intersection of lines $3\mathrm{x}-4\mathrm{y}+1=0$ and $8\mathrm{x}+6\mathrm{y}+1=0$ is $\left(\frac{-1}{5},\frac{1}{10}\right)$
Now, $\sum _{\mathrm{i}=1}^4{\mathrm{x}}_i=4\left(-\frac{1}{5}\right)=-\frac{4}{5}$
and $\sum _{\mathrm{i}=1}^4{\mathrm{y}}_i=4\left(\frac{1}{10}\right)=\frac{2}{5}$
$\Rightarrow \frac{\sum _{\mathrm{i}=1}^4{\mathrm{x}}_{\mathrm{i}}}{\sum _{\mathrm{i}=1}^4{\mathrm{y}}_{\mathrm{i}}}=\frac{-\frac{4}{5}}{\frac{2}{5}}=-2$

Hence, the correct answer is option: 'B'.