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
Correct Answer: C
Solution :
Since, the point A (2, 1) is translated parallel to \[x-y=3,\] therefore AA' has the same slope as that of \[x-y=3.\] |
Therefore, AA' passes through (2, 1) and has the slope of 1. |
Here, \[\tan \theta =1\] |
\[\Rightarrow \]\[\cos \theta =1\sqrt{2},\]\[\sin \theta =1/\sqrt{2}\] |
Thus, the equation of AA' is |
\[\frac{x-2}{\cos \,\,(\pi /4)}=\frac{y-1}{\sin \,\,(\pi /4)}\] |
Since, AA'= 4, therefore the coordinates of A' are given by |
\[\frac{x-2}{\cos \,\,(\pi /4)}=\frac{y-1}{\sin \,\,(\pi /4)}=-\,\,4\] |
\[\Rightarrow \] \[x=2-4\cos \frac{\pi }{4},\]\[y=1-4\sin \frac{\pi }{4}\] |
\[\Rightarrow \] \[x=2-2\sqrt{2},\]\[y=1-2\sqrt{2}\] |
Hence, the coordinates of A' are |
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