Enlarge shape A by scale factor \( \frac{1}{3} \) centre (0, 1) - Edexcel - GCSE Maths - Question 8 - 2018 - Paper 2

To enlarge the shape by a scale factor of ( \frac{1}{3} ) about the centre (0, 1), we use the formula for enlargement:

[ (x', y') = (0 + k(x-0), 1 + k(y-1)) ]

where ( k ) is the scale factor. Substituting ( k = \frac{1}{3} ), we have:

[ (x', y') = \left( \frac{1}{3}(x), \frac{1}{3}(y-1) + 1 \right) ]

Now, we will calculate the new coordinates for each vertex:

1. For Vertex 1 (2, 8):
   ( x' = \frac{1}{3}(2) = \frac{2}{3} )
   ( y' = \frac{1}{3}(8-1) + 1 = \frac{7}{3} + 1 = \frac{10}{3} ) New Vertex 1: ( \left( \frac{2}{3}, \frac{10}{3} \right) )

2. For Vertex 2 (2, 6):
   ( x' = \frac{1}{3}(2) = \frac{2}{3} )
   ( y' = \frac{1}{3}(6-1) + 1 = \frac{5}{3} + 1 = \frac{8}{3} ) New Vertex 2: ( \left( \frac{2}{3}, \frac{8}{3} \right) )

3. For Vertex 3 (4, 6):
   ( x' = \frac{1}{3}(4) = \frac{4}{3} )
   ( y' = \frac{1}{3}(6-1) + 1 = \frac{5}{3} + 1 = \frac{8}{3} ) New Vertex 3: ( \left( \frac{4}{3}, \frac{8}{3} \right) )

4. For Vertex 4 (4, 8):
   ( x' = \frac{1}{3}(4) = \frac{4}{3} )
   ( y' = \frac{1}{3}(8-1) + 1 = \frac{7}{3} + 1 = \frac{10}{3} ) New Vertex 4: ( \left( \frac{4}{3}, \frac{10}{3} \right) )

The new vertices of the enlarged shape A are as follows:

• New Vertex 1: ( \left( \frac{2}{3}, \frac{10}{3} \right) )
• New Vertex 2: ( \left( \frac{2}{3}, \frac{8}{3} \right) )
• New Vertex 3: ( \left( \frac{4}{3}, \frac{8}{3} \right) )
• New Vertex 4: ( \left( \frac{4}{3}, \frac{10}{3} \right) )