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 $A^{\prime }$ is in the third quadrant, then the coordinates of $A^{\prime }$ 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

Answer 1

Answer: (C)

Since the point $A(2,1)$ is translated parallel to $x-y=3,A{A}^{\prime }$
has the same slope as that of $x-y=3$. Therefore, $A{A}^{\prime }$ 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 $A{A}^{\prime }=4$, the coordinates of $A^{\prime }$ are given by
$\frac{x-2}{cos(\pi /4)}=\frac{y-1}{sin(\pi /4)}=-4$
or $x=2-4cos\frac{\pi }{4},y=1-4sin\frac{\pi }{4}$
or $x=2-2\sqrt{2},y=1-2\sqrt{2}$
Hence, the coordinates of $A^{\prime }$ are $(2-2\sqrt{2},1-2\sqrt{2})$.