Transformations: Question 2

Worked solution

Key idea

For an enlargement with centre $O$ and scale factor $k$, every point $P$ maps to an image $P'$ given by

$P' = O + k\,(P - O).$

In words: find the vector from the centre to the point, multiply it by $k$, then measure that new vector from the centre. A negative $k$ sends the image to the opposite side of the centre and turns the shape upside down; a fractional $|k|<1$ makes the image smaller.

Here $O = (1, 2)$ and $k = -\tfrac{1}{2}$.

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Part (a): image coordinates

Work out each vertex in turn.

Vertex $A(5,4)$: $A - O = (5-1,\; 4-2) = (4, 2)$ $k(A-O) = -\tfrac{1}{2}(4, 2) = (-2, -1)$ $A' = O + (-2,-1) = (1-2,\; 2-1) = (-1, 1).$

Vertex $B(9,4)$: $B - O = (8, 2), \qquad -\tfrac{1}{2}(8,2) = (-4,-1)$ $B' = (1-4,\; 2-1) = (-3, 1).$

Vertex $C(5,10)$: $C - O = (4, 8), \qquad -\tfrac{1}{2}(4,8) = (-2,-4)$ $C' = (1-2,\; 2-4) = (-1, -2).$

$\boxed{A'(-1,\,1),\quad B'(-3,\,1),\quad C'(-1,\,-2)}$

Check (collinearity through the centre): the centre, each object point and its image should be collinear. For $A$: from $O(1,2)$ the vector to $A$ is $(4,2)$ and to $A'$ is $(-2,-1) = -\tfrac12(4,2)$: same line, opposite direction, half the length. Good. The image is on the opposite side of $O$ and half the size, exactly as a factor of $-\tfrac12$ requires.

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Part (b): the reverse transformation

The map in part (a) is an enlargement. The transformation that undoes an enlargement is another enlargement about the same centre, with the reciprocal scale factor:

$\frac{1}{k} = \frac{1}{-\tfrac{1}{2}} = -2.$

So $A'B'C' \to F$ is an enlargement, centre $(1,2)$, scale factor $-2$.

Verification with $A'(-1,1)$: $O + (-2)\big(A' - O\big) = (1,2) + (-2)\big((-1-1),\,(1-2)\big) = (1,2) + (-2)(-2,-1) = (1,2)+(4,2) = (5,4) = A.\ \checkmark$

$\boxed{\text{Enlargement, centre }(1,2),\ \text{scale factor } -2}$

A common slip is to call this a $180^\circ$ rotation. A rotation would keep the shape the same size; here the image $F$ is twice the size of $A'B'C'$, so it must be an enlargement (the negative factor produces the half-turn appearance, but the size change rules out a rotation).

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Part (c): area

Lengths scale by the (magnitude of the) linear scale factor, so areas scale by its square:

$\text{area scale factor} = k^2 = \left(-\tfrac{1}{2}\right)^2 = \tfrac{1}{4}.$

(The sign disappears on squaring; a negative scale factor never gives a negative area.)

$\text{Area of } A'B'C' = 12 \times \tfrac{1}{4} = 3 \text{ square units}.$

Independent check from the coordinates: $A'(-1,1)$ and $B'(-3,1)$ give a horizontal base of length $|-1-(-3)| = 2$; the perpendicular height up to $C'(-1,-2)$ is $|1-(-2)| = 3$. Then $\text{area} = \tfrac{1}{2}\times 2 \times 3 = 3 \text{ square units}. \checkmark$

$\boxed{\text{Area scale factor } = \tfrac{1}{4}, \qquad \text{Area of } A'B'C' = 3 \text{ square units}}$