Consider a triangle A(0,0), B(1,1) and C(5,2). The triangle has to be rotated by an angle of 45° about the point P(-1, -1). What shall be the coordinates of the new triangle ?
\(A' = \left(-1, \sqrt{2} - 1\right), B' = \left(-1, 2\sqrt{2} - 1\right)\) and \(C' = \left(\frac{3}{2}\sqrt{2} - 1, \frac{9}{2}\sqrt{2} - 1\right)\)
Rotation by \(\theta\) about \(P(x_{p}, y_{p})\): \(x' = x_{p} + (x - x_{p})\cos\theta - (y - y_{p})\sin\theta\) and \(y' = y_{p} + (x - x_{p})\sin\theta + (y - y_{p})\cos\theta\), with \(\cos 45^{\circ} = \sin 45^{\circ} = \frac{1}{\sqrt{2}}\).
A: relative point \((1, 1)\), so \(x' = -1 + \frac{1 - 1}{\sqrt{2}} = -1\) and \(y' = -1 + \frac{1 + 1}{\sqrt{2}} = \sqrt{2} - 1\)
B: relative point \((2, 2)\), so \(x' = -1 + 0 = -1\) and \(y' = -1 + \frac{4}{\sqrt{2}} = 2\sqrt{2} - 1\)
C: relative point \((6, 3)\), so \(x' = -1 + \frac{3}{\sqrt{2}} = \frac{3}{2}\sqrt{2} - 1\) and \(y' = -1 + \frac{9}{\sqrt{2}} = \frac{9}{2}\sqrt{2} - 1\)
Check: distances from P are kept, e.g. \(|PA| = \sqrt{2}\) and \(|PA'| = \sqrt{0^{2} + (\sqrt{2})^{2}} = \sqrt{2}\). The other choices place A' or B' at points not at the rotated positions.
Hence, \(A' = \left(-1, \sqrt{2} - 1\right), B' = \left(-1, 2\sqrt{2} - 1\right)\) and \(C' = \left(\frac{3}{2}\sqrt{2} - 1, \frac{9}{2}\sqrt{2} - 1\right)\).
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