Question

The origin is translated to $(1,2)$. The point $(7,5)$ in the old system undergoes the following transformation successively.
(i) Moves to the new point under the given translation of the origin.
(ii)Translated through $2$ units along the negative direction of the new $x$-axis
(iii) Rotated through an angle $\frac{\pi }{4}$ about the origin of the new system in the clockwise direction.

The final position of the point $(7,5)$ is

• A. $\left(\frac{9}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)$
• B. $\left(\frac{7}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
• C. $\left(\frac{7}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)$
• D. $\left(\frac{5}{\sqrt{2}},\frac{-1}{\sqrt{2}}\right)$

Answer 1

Answer: (C)

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

i) After translation of origin, new position $P_1=(7-1,5-2)=(6,3)$

ii) After translating $2$ units in -ve dis. new position $P_2=(6-2,3)=(4,3)$

iii) convert it to polar points

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

$\tan Q=\frac{3}{4}$ $\cos Q=\frac{4}{5}$ $\sin Q=\frac{3}{5}$

Now after rotation by $\frac{\pi }{4}$, final co-ordinates $P_3=$

$P_3=\left(5\cos \right(Q-\pi /4),5\sin (Q-\pi /4\left)\right)$

$=(5\times \frac{4}{5}\cdot \frac{1}{\sqrt{2}}+5\cdot \frac{3}{\sqrt{5}}\cdot \frac{1}{\sqrt{2}},5\cdot \frac{3}{5}\frac{1}{\sqrt{2}}-5\cdot \frac{4}{5}\cdot \frac{1}{\sqrt{2}})$

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