question_answer

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 third quadrant, then the coordinates of A' are

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