Question

The point A(4, 1) undergoes following transformations successively
(i) reflection about line y = x
(ii) translation through a distance of 2 units in the positive direction of x axis
(iii) rotation through an angle $\pi /4$ in anti clockwise direction about origin O
Then the final position of point A is

• A. $(\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$
• B. $(-2,7\sqrt{2})$
• C. $(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$
• D. none of these

Answer 1

The correct option is C $(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$

Given Point $A(4,1)$

$\left(i\right)$ reflection about $y=x$

Hence putting coordinates of point A in given eq $y=4,x=1$

The point becomes $A(1,4)$

$\left(ii\right)$ translation through distance of 2 units in positive X axis

$A(1+2,4)\Rightarrow A(3,4)$

$\left(iii\right)$ rotation of point through and angle $\frac{\pi }{4}$ in anticlockwise about origin O

After rotation Point will be $A\left(r\cos \right(\alpha +\frac{pi}{4}),r\sin (\alpha +\frac{pi}{4}\left)\right)$

Converting into polar form

$r=\sqrt{3^2+4^2}=5$

$\tan \alpha =\frac{4}{3}$

Hence $\cos \alpha =\frac{3}{5}$

$\sin \alpha =\frac{4}{5}$

$\cos (\alpha +\frac{\pi }{4})=\cos \alpha \cos \frac{\pi }{4}-\sin \alpha \sin \frac{\pi }{4}$

$\cos (\alpha +\frac{\pi }{4})=\frac{3}{5}\frac{1}{\sqrt{2}}-\frac{4}{5}\frac{1}{\sqrt{2}}$

$\cos (\alpha +\frac{\pi }{4})=\frac{-1}{5\sqrt{2}}$

$\sin (\alpha +\frac{\pi }{4})=\sin \alpha \cos \frac{\pi }{4}+\cos \alpha \sin \frac{\pi }{4}$

$\sin (\alpha +\frac{\pi }{4})=\frac{4}{5}\frac{1}{\sqrt{2}}+\frac{3}{5}\frac{1}{\sqrt{2}}$

$\sin (\alpha +\frac{\pi }{4})=\frac{7}{5\sqrt{2}}$

$A(5\times \frac{-1}{5\sqrt{2}},5\times \frac{7}{5\sqrt{2}})$

$A(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$