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 $\frac{\sum _{i=1}^4{x}_i}{\sum _{i=1}^4{y}_i}$ 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, $\sum _{i=1}^4{x}_i=4\left(-\frac{1}{5}\right)=-\frac{4}{5}$
and $\sum _{i=1}^4{y}_i=4\left(\frac{1}{10}\right)=\frac{2}{5}$
$\Rightarrow \frac{\sum _{i=1}^4{x}_i}{\sum _{i=1}^4{y}_i}=\frac{-\frac{4}{5}}{\frac{2}{5}}=-2$