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
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Answer:
Correct Answer: B
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 $ \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\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
