Rotation of Axes About Origin - Practice Questions & MCQ

Rotation of Axes About Origin  

$P(x, y)$ is the point in the original coordinate system and axes are rotated by an angle $\ominus$ anticlockwise direction about the origin. Then, the coordinates of point P with respect to the new coordinate system is $(\mathrm{X}, \mathrm{Y})=(\mathrm{x} \cos \theta+\mathrm{y} \sin \theta, \mathrm{y} \cos \theta-\mathrm{x} \sin \theta)$.

OX and OY are original system of coordinate axes and OX' and OY' are the new system of coordinate axes. PM and PN are perpendicular to OX and OX' and also NL and NQ perpendicular OX and PM .

We have

From the figure:

$
\mathrm{OM}=\mathrm{x}, \mathrm{PM}=\mathrm{y} \mathrm{ON}=\mathrm{X} \text { and } \mathrm{PN}=\mathrm{Y}
$
Now,

$
\mathrm{x}=\mathrm{OM}=\mathrm{OL}-\mathrm{ML}
$

$\because$ angle between two lines $=$ angles between their perpendiculars

$
\begin{aligned}
& =\mathrm{OL}-\mathrm{QN}=\mathrm{ON} \cos \theta-\mathrm{PN} \sin \theta \\
& =\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta
\end{aligned}
$

i.e. $\mathbf{x}=\mathbf{X} \cos \theta-\mathbf{Y} \sin \theta$

And,

$
\begin{aligned}
\mathrm{y} & =\mathrm{PM}=\mathrm{PQ}+\mathrm{QM}=\mathrm{PQ}+\mathrm{NL} \\
& =\mathrm{PN} \cos \theta+\mathrm{ON} \sin \theta \\
& =\mathrm{Y} \cos \theta+\mathrm{X} \sin \theta
\end{aligned}
$

i.e. $\mathbf{y}=\mathbf{Y} \cos \theta+\mathbf{X} \sin \theta$

Solving (i) and (ii), we get

$
\begin{aligned}
& X=x \cos \theta+y \sin \theta \\
& Y=y \cos \theta-x \sin \theta
\end{aligned}
$

AID TO MEMORY

\begin{array}{|c|c|c|}
\hline & x & y \\
\hline X & \cos \theta & \sin \theta \\
\hline Y & -\sin \theta & \cos \theta \\
\hline
\end{array}