First slide

Cartesian plane

Question

The point A (2, l) is translated parallel to the line x - y = 3 by a distance of 4 units. If the new position A¹ is in the third quadrant, then the coordinates of A¹are

Moderate

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

Solution

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) $and has slope 1. Here, \tan θ = 1 or \cos θ = 1 / \sqrt{2}, \sin θ = 1 / \sqrt{2}$
Thus, the equation of AA' is $\frac{x - 2}{\cos (π / 4)} = \frac{y - 1}{\sin (π / 4)}$
Since AA' = 4, the coordinates of A' are given by $\frac{x - 2}{\cos (π / 4)} = \frac{y - 1}{\sin (π / 4)} = - 4$
$\begin{aligned} or x = 2 - 4 \cos \frac{π}{4}, y = 1 - 4 \sin \frac{π}{4} \\ or x = 2 - 2 \sqrt{2}, y = 1 - 2 \sqrt{2} \end{aligned}$
$Hence, the coordinates of A^{'} are (2 - 2 \sqrt{2}, 1 - 2 \sqrt{2}) .$

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