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} $