Consider a triangle represented by A(0, 0), B(1, 1), C(5, 2). The triangle is rotated by 45 degrees about a point P(–1, –1). The co-ordinates of the new triangle obtained after rotation shall be _______
\(A'\left(-1,\ \sqrt{2}-1\right),\ B'\left(-1,\ 2\sqrt{2}-1\right),\ C'\left(\frac{3}{2}\sqrt{2}-1,\ \frac{9}{2}\sqrt{2}-1\right)\)
Rotation about an arbitrary pivot P(h, k) is a composite transformation: translate P to the origin, rotate about the origin, then translate back. For an anticlockwise rotation by θ: \(x' = h + (x-h)\cos\theta - (y-k)\sin\theta,\ \ y' = k + (x-h)\sin\theta + (y-k)\cos\theta\).
Here h = k = –1 and θ = 45°, so \(\cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}}\). With \(d_x = x+1,\ d_y = y+1\): \(x' = -1 + \frac{d_x - d_y}{\sqrt{2}},\ \ y' = -1 + \frac{d_x + d_y}{\sqrt{2}}\).
A(0, 0): \(d_x = d_y = 1\), so \(x' = -1 + 0 = -1\) and \(y' = -1 + \frac{2}{\sqrt{2}} = \sqrt{2}-1\). Thus \(A'(-1,\ \sqrt{2}-1)\).
B(1, 1): \(d_x = d_y = 2\), so \(x' = -1\) and \(y' = -1 + \frac{4}{\sqrt{2}} = 2\sqrt{2}-1\). Thus \(B'(-1,\ 2\sqrt{2}-1)\).
C(5, 2): \(d_x = 6,\ d_y = 3\), so \(x' = -1 + \frac{3}{\sqrt{2}} = \frac{3}{2}\sqrt{2}-1\) and \(y' = -1 + \frac{9}{\sqrt{2}} = \frac{9}{2}\sqrt{2}-1\).
A' and B' must lie on the vertical line x = –1, because A and B lie on the 45° line through P and a 45° anticlockwise turn makes that line vertical. Placing A' or B' on y = –1, or swapping the coordinates of C', does not match this rotation.
Hence, the answer is \(A'\left(-1,\ \sqrt{2}-1\right),\ B'\left(-1,\ 2\sqrt{2}-1\right),\ C'\left(\frac{3}{2}\sqrt{2}-1,\ \frac{9}{2}\sqrt{2}-1\right)\).
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