Slide, flip, turn, scale
Tonight's idea
Put a shape on a coordinate grid and every move becomes a rule for the coordinates. The arrow \(\to\) means "goes to".
- Translation (slide) by \(a\) across and \(b\) up: \((x, y) \to (x + a,\ y + b)\).
- Reflection (flip) in the \(x\)-axis: \((x, y) \to (x, -y)\). In the \(y\)-axis: \((x, y) \to (-x, y)\). In the line \(y = x\): \((x, y) \to (y, x)\).
- Rotation (turn) about the origin by 90° anticlockwise: \((x, y) \to (-y, x)\). By 180°: \((x, y) \to (-x, -y)\).
- Dilation (scale) from the origin by factor \(k\): \((x, y) \to (kx, ky)\).
Why does the 90° rule work? The point \((x, y)\) is the far corner of a rectangle with sides \(x\) and \(y\) along the axes. Give the rectangle a quarter turn: the side of length \(y\) now runs along the negative \(x\)-axis and the side of length \(x\) runs up the \(y\)-axis. The corner lands at \((-y, x)\).
The first three moves are rigid motions: every length and every angle is unchanged, so the image is congruent to the original, the same shape in a new position. Dilation by \(k \ne 0\) multiplies every length by \(|k|\) and keeps every angle, so the image is similar: the same shape at a different size. Its area is multiplied by \(k^2\), because width and height both scale by \(|k|\). (A negative \(k\) adds a half turn; \(k = 0\) squashes everything to one point.)
Watch one: reflect, then scale
Triangle \(P(2,1)\), \(Q(5,1)\), \(R(2,3)\). Reflect in the line \(y = x\) by swapping coordinates: \(P'(1,2)\), \(Q'(1,5)\), \(R'(3,2)\). Check a length: \(PQ = 3\) along the horizontal, and \(P'Q'\) runs from \((1,2)\) to \((1,5)\), also 3. Congruent, as a reflection should be. Now dilate by \(k = 2\): \((2,4)\), \((2,10)\), \((6,4)\). The side that was 3 is now 6, and the area goes from \(\tfrac12 \times 3 \times 2 = 3\) to \(\tfrac12 \times 6 \times 4 = 12\), four times as much.
Press the buttons and watch the coordinates change: mathdetours.com/demos/rigid.html. Later these same moves become matrices: mathdetours.com/demos/matrix.html.
Your turn
- Triangle \(A(1,2)\), \(B(4,2)\), \(C(4,6)\). Write the coordinates of the image under (a) a translation by 3 left and 1 up, (b) a reflection in the \(x\)-axis, (c) a rotation of 180° about the origin.
- Rotate the same triangle 90° anticlockwise about the origin. Then check that it is congruent: find \(AB\) and \(A'B'\), and find \(AC\) and \(A'C'\) using Pythagoras.
- Dilate the same triangle from the origin by \(k = \tfrac12\). Write the image coordinates, then find the area of the original and the area of the image. What is the ratio?
- Start from the point \((3, 5)\). (a) Reflect in the \(y\)-axis, then reflect the result in the \(x\)-axis. Which single rule from the list does the same job? (b) Reflect in the \(x\)-axis, then in the line \(y = x\). Compare with a 90° anticlockwise rotation. (c) Now do (b) in the other order: \(y = x\) first, then the \(x\)-axis. Is the result the same?
Stretch: turning about a different point
Rotate \((5, 3)\) by 90° anticlockwise about the point \((2, 1)\), not the origin. Method: slide everything so that \((2, 1)\) lands on the origin, rotate with the rule, slide back. Then write the rule for a general point \((x, y)\) rotated 90° anticlockwise about \((2, 1)\).
Answers
- (a) \((-2,3)\), \((1,3)\), \((1,7)\). (b) \((1,-2)\), \((4,-2)\), \((4,-6)\). (c) \((-1,-2)\), \((-4,-2)\), \((-4,-6)\).
- \(A'(-2,1)\), \(B'(-2,4)\), \(C'(-6,4)\). \(AB = 3\) and \(A'B'\) is the vertical gap from 1 to 4, also 3. \(AC = \sqrt{3^2 + 4^2} = 5\) and \(A'C' = \sqrt{4^2 + 3^2} = 5\). Same lengths, so congruent.
- \((\tfrac12, 1)\), \((2, 1)\), \((2, 3)\). Original area \(\tfrac12 \times 3 \times 4 = 6\); image \(\tfrac12 \times 1.5 \times 2 = 1.5\). Ratio \(\tfrac14 = k^2\).
- (a) \((3,5) \to (-3,5) \to (-3,-5)\), the same as a 180° rotation. Two reflections in perpendicular lines make a half turn. (b) \((3,5) \to (3,-5) \to (-5,3)\), exactly the 90° rotation. (c) \((3,5) \to (5,3) \to (5,-3)\): a 90° turn the other way. The order matters.
Stretch. Slide by \((-2,-1)\): \((5,3) \to (3,2)\). Rotate: \((3,2) \to (-2,3)\). Slide back by \((2,1)\): \((0,4)\). In general \((x, y) \to (x-2,\ y-1) \to (1-y,\ x-2) \to (3-y,\ x-1)\). Check: \((5,3) \to (0,4)\), and the centre \((2,1) \to (2,1)\) stays put, as it must.