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

• 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 $x - y = 3, AA'$
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 - 4\,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 $(2 - 2\sqrt{2}, 1 - 2\sqrt{2})$.