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If the axes are rotated through an angle \(45^{\circ}\), then the co-ordinates of the point \((4 \sqrt{2},-6 \sqrt{2})\) in the new system are ____
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Verified Answer
The correct answer is:
\((-2,-10)\)
Let \((x, y)\) are coordinates related to old axes and \((X, Y)\) are related to rotated (new) axes then,
\(\begin{gathered}
x=X \cos \theta-Y \sin \theta \\
y=X \sin \theta+Y \cos \theta
\end{gathered}\)
Here,
\(\begin{aligned}
& \theta=45^{\circ} \\
& x=4 \sqrt{2}, y=-6 \sqrt{2}
\end{aligned}\)
So, \(\quad 4 \sqrt{2}=\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\) and \(-6 \sqrt{2}=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}\)
Adding we have,
\(-2 \sqrt{2}=2 \cdot \frac{X}{\sqrt{2}} \Rightarrow X=-2\)
and subtracting,
\(10 \sqrt{2}=-\frac{2 Y}{\sqrt{2}} \Rightarrow Y=-10 .\)
\(\begin{gathered}
x=X \cos \theta-Y \sin \theta \\
y=X \sin \theta+Y \cos \theta
\end{gathered}\)
Here,
\(\begin{aligned}
& \theta=45^{\circ} \\
& x=4 \sqrt{2}, y=-6 \sqrt{2}
\end{aligned}\)
So, \(\quad 4 \sqrt{2}=\frac{X}{\sqrt{2}}-\frac{Y}{\sqrt{2}}\) and \(-6 \sqrt{2}=\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}\)
Adding we have,
\(-2 \sqrt{2}=2 \cdot \frac{X}{\sqrt{2}} \Rightarrow X=-2\)
and subtracting,
\(10 \sqrt{2}=-\frac{2 Y}{\sqrt{2}} \Rightarrow Y=-10 .\)
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