4.2 Q-3

Question Statement

The problem involves rotating the coordinate axes about the origin by a specified angle and finding the new coordinates of a point in the rotated system. The task is to apply this transformation to several given points.

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Background and Explanation

To solve this, we need to understand the concept of rotating coordinate axes. When the axes are rotated by an angle $\theta$, the coordinates of a point change according to the following transformation formulas:

• For the new X-coordinate:

$$X = x \cos \theta + y \sin \theta$$

• For the new Y-coordinate:

$$Y = -x \sin \theta + y \cos \theta$$

Where $(x, y)$ are the coordinates in the original system and $(X, Y)$ are the coordinates in the rotated system.

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Solution

i. $\mathbf{P(5, 3)}, \theta = 45^\circ$

We are given the point $P(5, 3)$ and the rotation angle $\theta = 45^\circ$. We can use the transformation formulas to find the new coordinates.

1. Calculate X-coordinate:

$$X = 5 \cos 45^\circ + 3 \sin 45^\circ = \frac{5}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{8}{\sqrt{2}} = 4 \sqrt{2}$$

2. Calculate Y-coordinate:

$$Y = -5 \sin 45^\circ + 3 \cos 45^\circ = \frac{-5}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{-2}{\sqrt{2}} = -\sqrt{2}$$

Thus, the new coordinates are $P(4 \sqrt{2}, -\sqrt{2})$.

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ii. $\mathbf{P(3, -7)}, \theta = 30^\circ$

Now, we apply the transformation to the point $P(3, -7)$ with $\theta = 30^\circ$.

1. Calculate X-coordinate:

$$X = 3 \cos 30^\circ + (-7) \sin 30^\circ = 3 \left( \frac{\sqrt{3}}{2} \right) - 7 \left( \frac{1}{2} \right) = \frac{3\sqrt{3} - 7}{2}$$

2. Calculate Y-coordinate:

$$Y = -3 \sin 30^\circ + (-7) \cos 30^\circ = -3 \left( \frac{1}{2} \right) - 7 \left( \frac{\sqrt{3}}{2} \right) = \frac{-7 - 3\sqrt{3}}{2}$$

Thus, the new coordinates are $P\left( \frac{3\sqrt{3} - 7}{2}, \frac{-7 - 3\sqrt{3}}{2} \right)$.

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iii. $\mathbf{P(11, -15)}, \theta = 60^\circ$

For this point, we have $P(11, -15)$ and $\theta = 60^\circ$. Let’s apply the transformation.

1. Calculate X-coordinate:

$$X = 11 \cos 60^\circ + (-15) \sin 60^\circ = 11 \left( \frac{1}{2} \right) - 15 \left( \frac{\sqrt{3}}{2} \right) = \frac{11 - 15\sqrt{3}}{2}$$

2. Calculate Y-coordinate:

$$Y = -11 \sin 60^\circ + (-15) \cos 60^\circ = -11 \left( \frac{\sqrt{3}}{2} \right) - 15 \left( \frac{1}{2} \right) = \frac{-11 - 15\sqrt{3}}{2}$$

Thus, the new coordinates are $P\left( \frac{11 - 15\sqrt{3}}{2}, \frac{-11 - 15\sqrt{3}}{2} \right)$.

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iv. $\mathbf{P(15, 10)}, \theta = \arctan\left(\frac{1}{2}\right)$

Finally, for the point $P(15, 10)$ and $\theta = \arctan\left(\frac{1}{2}\right)$, we follow the same procedure.

1. Find $\cos \theta$ and $\sin \theta$: Using the identity $\tan \theta = \frac{1}{2}$, we find:

$$\sin \theta = \frac{1}{\sqrt{10}}, \quad \cos \theta = \frac{3}{\sqrt{10}}$$

2. Calculate X-coordinate:

$$X = 15 \cos \theta + 10 \sin \theta = 15 \left( \frac{3}{\sqrt{10}} \right) + 10 \left( \frac{1}{\sqrt{10}} \right) = \frac{45 + 10}{\sqrt{10}} = \frac{55}{\sqrt{10}}$$

3. Calculate Y-coordinate:

$$Y = 15 \sin \theta + 10 \cos \theta = 15 \left( \frac{1}{\sqrt{10}} \right) + 10 \left( \frac{3}{\sqrt{10}} \right) = \frac{15 + 30}{\sqrt{10}} = \frac{45}{\sqrt{10}}$$

Thus, the new coordinates are $P\left( \frac{55}{\sqrt{10}}, \frac{45}{\sqrt{10}} \right)$.

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Key Formulas or Methods Used

• Transformation for X-coordinate:

$$X = x \cos \theta + y \sin \theta$$

• Transformation for Y-coordinate:

$$Y = -x \sin \theta + y \cos \theta$$

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Summary of Steps

1. Identify the point’s original coordinates $(x, y)$ and the angle of rotation $\theta$.
2. Apply the transformation formulas to calculate the new X and Y coordinates:
   • $X = x \cos \theta + y \sin \theta$
   • $Y = -x \sin \theta + y \cos \theta$
3. Simplify the expressions to find the new coordinates $(X, Y)$.

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