Question

The distance of the point $(1,2)$ from the line $x+y+5=0$ measured along the line parallel to $3x-y=7$ is equal to

• A. $4\sqrt{10}$
• B. $40$
• C. $\sqrt{40}$
• D. $10\sqrt{2}$
• E. $2\sqrt{20}$

Answer 1

Answer: (C)

Let equation of line parallel to $3x-y=7$ be $3x-y=\lambda$ .
It passes through (1, 2).
$\therefore$ $3-2=\lambda \Rightarrow \lambda =1$
$\therefore$ Line is $3x-y=1$
The point of intersection of
$x+y+5=0$ and $3x-y-1$ is $(-1,-4)$ .
$\therefore$ Distance between (1, 2) and $(-1,-4)$
$=\sqrt{{(2)}^2+{(6)}^2}=\sqrt{40}$