Interpret and use fractional {and negative} scale factors for enlargements
Here we consider enlargements with fractional scale factors. Some of the tasks are (or may appear) more challenging due to the fact that the centre of enlargement lies outside the confines of the diagram. The idea here is to encourage students to think about the properties of enlargement and perhaps be more analytical. Of course, they (or their teacher) can, if they wish, extend the diagram to include the centre.
MONDAY: Here the centre of enlargement lies slightly below and to the left of the given grid. However, a short pause for thought might enable students to see that the desired image can be drawn without recourse to the centre.
TUESDAY: Here we can again find the desired image without finding/using the centre of enlargement. However, the centre is within the confines of the given diagram this time and some students might find it helpful to find it.
The length of A'D' is ⅔ the length of AD, which means that the scale factor is ×⅔ again. In turn, this means that C'D' = ⅔ × CD = ⅔ × 9 = 6 units. So we can now complete the image of the rectangle. However, even if we do this without first finding the centre of enlargement, it is quite nice to confirm that the lines through A and A', B and B', C and C' and D and D' all meet at a point (P, say). It is also nice to confirm that PA' = ⅔ × PA, PB' = ⅔ × PB, etc.
Note: It is highly likely that some students will complete the rectangle like this. The result looks plausible but the rectangles are not similar. One way to show this is to draw lines through the corresponding vertices A and A', B and B', etc. The lines do not all meet at a single point (the putative centre of enlargement).
WEDNESDAY: Here the 6 by 9 rectangle is 'enlarged' to become a 2 by 3 rectangle. So we can draw the two desired images without needing to find the centres. However, it is conceptually helpful to find the centres, to tie things together.
THURSDAY: Here the two desired centres are some way off the page, and it might well be useful to extend the diagram at some stage, if only to check/confirm students' answers.
1.5 × CA = CA'.
For part ii. the relation (for the new C) is ⅔ × CA = CA', which leads to the finding that C now has coordinates (22, 15). We can show the centres by extending the given diagram. Notice the symmetrical nature of the centres' positions. Will that always happen for pairs of scale factors a/b and b/a?
FRIDAY: To find the desired scale factor in this task, we need to compare corresponding lengths on the object and image. The easiest to read are the heights, which look to be 2 and 5 units long. This suggests the scale factor is ×2½.
In part b), it would be quite challenging to locate the centre of enlargement, which anyway lies outside the confines of the given diagram. However, we can answer part b) without knowing where the centre is. Moving the centre 2 cm to the right means the object is a horizontal distance of 2 cm further from the centre, so the image will be a horizontal distance of 2½ × 2 cm = 5 cm further from the centre's new position. In other words, the image moves 5 cm – 2 cm = 3 cm to the left of its original position.
In the diagram below, we have assumed the grid-lines are 1 cm apart. As can be seen, moving the centre of enlargement 2 cm to the right, and keeping to a scale factor of ×2½, moves the image 3 cm to the left.
