Q96 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics
Find the coordinates of $M$ in the original system if the point $M$ changes to $(4,-3)$ when the axes are rotated through an angle of $135^{\circ}$.
- A$\left(\frac{-1}{2}, \frac{7}{2}\right)$
- B$\left(\frac{1}{2}, \frac{7}{2}\right)$
- Correct answerC$\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
- D$\left(\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$
Solution
$$
\begin{aligned}
(*) \text { Let } M^{\prime}=(4,-3) & =\left(x^{\prime}, y^{\prime}\right) \\
\theta & =135^{\circ}
\end{aligned}
$$
Let coordinates of $M$ be $(x, y)$
we have,
$$
\begin{aligned}
\quad x & =x^{\prime} \cos \theta-y^{\prime} \sin \theta \text { and } y=x^{\prime} \sin \theta+y^{\prime} \cos \theta \\
\Rightarrow \quad x & =4 \cdot \cos 135+3 \sin 135^{\circ} \text { and } \\
y & =4 \cdot \sin 135^{\circ}+(-3) \cos 135^{\circ} \\
\Rightarrow \quad x & =4 \cdot\left(\frac{-1}{\sqrt{2}}\right)+3\left(\frac{1}{\sqrt{2}}\right) \text { and } \\
y & =4 \cdot \frac{1}{\sqrt{2}}+(-3) \frac{1}{\sqrt{2}} \\
\quad x & =-\frac{1}{\sqrt{2}} \text { and } y=\frac{1}{\sqrt{2}} \\
\therefore \quad M & =(x, y)=\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)
\end{aligned}
$$
(*) No option is correct.
Hence, the correct answer is option: 'C'.