If the coordinates of a point P are transformed to (4,-3) when the axes are rotated through an angle of `135^(0)`, then find P.

Answer

To find the original coordinates of point P before the axes were rotated, we will use the transformation formulas for coordinates when the axes are rotated by an angle θ.

Given that the new coordinates after rotation are (4, -3) and the angle of rotation θ is 135°, we can denote the original coordinates as (x, y) and the new coordinates as (x', y') = (4, -3).

### Step 1: Write the transformation formulas
The transformation formulas for rotating the axes are:
\[ x = x' \cos \theta - y' \sin \theta \]
\[ y = x' \sin \theta + y' \cos \theta \]

### Step 2: Substitute the known values
Here, \(x' = 4\), \(y' = -3\), and \(\theta = 135^\circ\). We need to calculate \(\cos 135^\circ\) and \(\sin 135^\circ\).

### Step 3: Calculate \(\cos 135^\circ\) and \(\sin 135^\circ\)
Using trigonometric identities:
- \(\cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ = -\frac{1}{\sqrt{2}}\)
- \(\sin 135^\circ = \sin(180^\circ - 45^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}}\)

### Step 4: Substitute the values into the equations
Now we can substitute these values into the transformation equations.

For \(x\):
\[ x = 4 \cdot \left(-\frac{1}{\sqrt{2}}\right) - (-3) \cdot \left(\frac{1}{\sqrt{2}}\right) \]
\[ x = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{-4 + 3}{\sqrt{2}} = \frac{-1}{\sqrt{2}} \]

For \(y\):
\[ y = 4 \cdot \left(\frac{1}{\sqrt{2}}\right) + (-3) \cdot \left(-\frac{1}{\sqrt{2}}\right) \]
\[ y = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{4 + 3}{\sqrt{2}} = \frac{7}{\sqrt{2}} \]

### Step 5: Write the final coordinates
Thus, the original coordinates of point P are:
\[ P = \left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right) \]

### Summary
The original coordinates of point P before the rotation are \(\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\).

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