Question

Let $\left(x_{1} , y_{1}\right),\left(x_{2} , y_{2}\right),\left(x_{3} , y_{3}\right)$ and $\left(x_{4} , y_{4}\right)$ are four points which are at unit distance from the lines $3x-4y+1=0$ and $8x+6y+1=0,$ then the value of $\frac{\displaystyle \sum _{i = 1}^{4} x_{i}}{\displaystyle \sum _{i = 1}^{4} y_{i}}$ is equal to

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

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