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 the 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: 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\[(2-2\sqrt{2},1-2\sqrt{2})\].