Question

The point $(4,1)$ undergoes the following three transformations successively
(a) Reflection about the line $y=x$

(b) Transformation through a distance $2$ units along the positive direction of the x-axis.

(c) Rotation through an angle $p/4$ about the origin in the anti clockwise direction.

The final position of the point is given by the co-ordinates

• A. $(\frac{4}{\sqrt{2}},\frac{1}{\sqrt{2}})$
• B. $(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$
• C. $(\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$
• D. $(-\frac{3}{\sqrt{2}},\frac{4}{\sqrt{2}})$

Answer 1

The correct option is B

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

Given Point $P(4,1)$

$\left(a\right)$ Reflection of given point about the line $y=x$

In Point P $x=4$ Hence about the line y-coordinate be $y=x\Rightarrow y=4$

Point P become $P(1,4)$

$\left(b\right)$ Transformation of Point P through distance $2$ units along the positive direction of the x-axis

$P(1+2,4)\equiv P(3,4)$

$\left(c\right)$Rotation of Point P through angle $\frac{\pi }{4}$ about origin in anti clockwise direction

$P(r\cos (\alpha +\frac{\pi }{4}),r\sin (\alpha +\frac{\pi }{4}))$

Given point $P(3,4)$

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

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

Hence In right angle triangle here hypotenuse be $5$

$\cos \alpha =\frac{3}{5}$ and $\sin \alpha =\frac{4}{5}$

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

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

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

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

$\sin (\alpha +\frac{\pi }{4})=\frac{4}{5}\times \frac{1}{\sqrt{2}}+\frac{3}{5}\times \frac{1}{\sqrt{2}}=\frac{7}{5\sqrt{2}}$

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

Point be

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