First slide

Cartesian plane

Question

The point (4, I ) undergoes the following three transformations successively.
i . Reflection about the line y: x.
ii. Translation through a distance 2 units along the positive direction of the x-axis.
iii. Rotation through an angle $\frac{π}{4}$ about the origin in the counterclockwise direction.
Then the final position of the point is given by the co ordinates

Moderate

A

$(1 / \sqrt{2}, 7 / \sqrt{2})$
B

$(- \sqrt{2}, 7 \sqrt{2})$
C

$(- 1 / \sqrt{2}, 7 / \sqrt{2})$
D

$(\sqrt{2}, 7 \sqrt{2})$

Solution

Reflection about the line y = x changes the point (4, l) to (1, 4).
On the translation of (1, 4) through a distance of 2 units along the positive direction of the x-axis, the point becomes (1 + 2. 4), i.e., (3,4).
On rotation about the origin through an angle $\frac{π}{4}$
point P takes the position P' such that $O P = O P^{'} . Also, O P = 5$
$= O P^{'} and \cos θ = 3 / 5, \sin θ = 4 / 5 . Now,$
$\begin{matrix} x = O P^{'} \cos (\frac{π}{4} + θ) \\ = 5 (\cos \frac{π}{4} \cos θ - \sin θ) \\ = 5 (\frac{3}{5 \sqrt{2}} - \frac{4}{5 \sqrt{2}}) \\ = - \frac{1}{\sqrt{2}} \\ y = O P^{'} \sin (\frac{π}{4} + θ) \\ = 5 (\sin \frac{π}{4} \cos θ + \cos \frac{π}{4} \sin θ) \\ = 5 (\frac{3}{5 \sqrt{2}} + \frac{4}{5 \sqrt{2}}) = \frac{7}{\sqrt{2}} \\ P^{'} \equiv (\frac{- 1}{\sqrt{2}}, \frac{7}{\sqrt{2}}) \end{matrix}$

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