Question: The point $ A(2,1) $ is translated parallel to the line $ x-y=3 $ by, a distance of 4 units. If the new position A? is in the third quadrant, then the coordinates of A? are
Options:
A) $ (2+2\sqrt{2},1+2\sqrt{2}) $
B) $ (-2+\sqrt{2},-1-2\sqrt{2}) $
C) $ (2-2\sqrt{2},1-2\sqrt{2}) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Since the point A(2, 1) is translated parallel to $ x-y=3, $ AA? has the same slope as that of $ x-y=3. $ Therefore, AA? passes through (2, 1) and has slope 1. Here, $ \tan \theta =1 $ or
Thus, the equation of AA? is
$ \cos \theta =1/\sqrt{2},\sin \theta =1/\sqrt{2} $
Thus, the equation of AA? is
$ \frac{x-2}{\cos (\pi /4)}=\frac{y-1}{\sin (\pi /4)} $
Since AA?=4, the coordinates of A? are given by
$ \frac{x-2}{\cos (\pi /4)}=\frac{y-1}{\sin (\pi /4)}=-4 $
or $ x=2-4\cos \frac{\pi }{4},y=1-4\sin \frac{\pi }{4} $
or $ x=2-2\sqrt{2},y=1-2\sqrt{2} $
Hence, the coordinates of A? are
$ (2-2\sqrt{2},1-2\sqrt{2}). $
BETA
