Transformations: Question 1

Syllabus C7.1, E7.1

Structured Core 7 marks

Triangle TT has vertices at (2,1)(2,1), (2,4)(2,4) and (4,1)(4,1).

Each triangle below is the image of triangle TT under a single transformation. Describe fully the single transformation that maps triangle TT onto each image.

(a) Triangle UU has vertices at (5,−4)(5,-4), (5,−1)(5,-1) and (7,−4)(7,-4). [2]

(b) Triangle VV has vertices at (1,2)(1,2), (4,2)(4,2) and (1,4)(1,4). [2]

(c) Triangle WW has vertices at (−1,2)(-1,2), (−4,2)(-4,2) and (−1,4)(-1,4). [3]

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Worked solution

For each part, first check the image is congruent to TT (a right-angled triangle with legs 33 and 22), then identify the one transformation and give every required detail.

Triangle TT: right angle at (2,1)(2,1), vertical leg (2,1)→(2,4)(2,1)\to(2,4) of length 33, horizontal leg (2,1)→(4,1)(2,1)\to(4,1) of length 22.

(a) Triangle UU: (5,−4),(5,−1),(7,−4)(5,-4),(5,-1),(7,-4)

The orientation and shape are unchanged, so this is a translation. The translation vector is found by subtracting an object point from its image point:

(5−4)−(21)=(3−5)\begin{pmatrix} 5 \\ -4 \end{pmatrix}-\begin{pmatrix} 2 \\ 1 \end{pmatrix}=\begin{pmatrix} 3 \\ -5 \end{pmatrix}

Check with another pair of vertices:

(7−4)−(41)=(3−5) ✓\begin{pmatrix} 7 \\ -4 \end{pmatrix}-\begin{pmatrix} 4 \\ 1 \end{pmatrix}=\begin{pmatrix} 3 \\ -5 \end{pmatrix}\ \checkmark

Description: a translation by the column vector (3−5)\boxed{\begin{pmatrix} 3 \\ -5 \end{pmatrix}}

(b) Triangle VV: (1,2),(4,2),(1,4)(1,2),(4,2),(1,4)

The orientation is reversed (the right angle moves from (2,1)(2,1) to (1,2)(1,2)), so this is a reflection. Match the vertices:

(2,1)→(1,2),(2,4)→(4,2),(4,1)→(1,4)(2,1)\to(1,2),\qquad (2,4)\to(4,2),\qquad (4,1)\to(1,4)

Each image swaps the xx- and yy-coordinates of its object point: (x,y)→(y,x)(x,y)\to(y,x). This is reflection in the line y=xy=x. Check the midpoint of (2,1)(2,1) and (1,2)(1,2) is (1.5,1.5)(1.5,1.5), which lies on y=xy=x. ✓

Description: a reflection in the line y=x\boxed{y=x}

(c) Triangle WW: (−1,2),(−4,2),(−1,4)(-1,2),(-4,2),(-1,4)

The shape is congruent and the orientation is preserved (not a reflection), but it is not a translation, so this is a rotation. Match the vertices:

(2,1)→(−1,2),(2,4)→(−4,2),(4,1)→(−1,4)(2,1)\to(-1,2),\qquad (2,4)\to(-4,2),\qquad (4,1)\to(-1,4)

Each image follows the rule (x,y)→(−y,x)(x,y)\to(-y,x), which is a quarter turn anticlockwise about the origin. For example (2,1)(2,1) in the first quadrant maps to (−1,2)(-1,2) in the second quadrant, confirming the anticlockwise direction. The centre (0,0)(0,0) is equidistant from each object–image pair, e.g. (2,1)(2,1) and (−1,2)(-1,2) are both a distance 5\sqrt{5} from the origin. ✓

A rotation must state the angle, the direction and the centre.

Description: a rotation through 90∘90^\circ anticlockwise, centre (0,0)\boxed{(0,0)}