The triangle ABC and the point D are shown on the co-ordinate diagram below - Junior Cycle Mathematics - Question 9 - 2016

The midpoint M of two points A(x1, y1) and B(x2, y2) is calculated as:

$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Using the coordinates for A and B: $M = \left( \frac{5 + 3}{2}, \frac{1 + 3}{2} \right) = \left( 4, 2 \right)$

The length of the line segment |AB| is calculated using the distance formula:

$|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Thus: $|AB| = \sqrt{(3 - 5)^2 + (3 - 1)^2}$ $= \sqrt{(-2)^2 + (2)^2}$ $= \sqrt{4 + 4}$ $= \sqrt{8} = 2\sqrt{2} \: \text{units}$

To find the area of triangle ABC, we can use the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Taking base AC and the height corresponding to point B, we find:

• Base AC = 4 units (from A(5, 1) to C(1, 1))
• Height = 3 – 1 = 2 units

Thus: $\text{Area} = \frac{1}{2} \times 4 \times 2 = 4 \: \text{square units}$

To perform central symmetry about the point D(4, 2), we need to reflect each vertex of triangle ABC across D:

• For A(5, 1):

  • X-coordinate = 4 - (5 - 4) = 3
  • Y-coordinate = 2 - (1 - 2) = 3
  • New coordinate A' = (3, 3)

• For B(3, 3):

  • X-coordinate = 4 - (3 - 4) = 5
  • Y-coordinate = 2 - (3 - 2) = 1
  • New coordinate B' = (5, 1)

• For C(1, 1):

  • X-coordinate = 4 - (1 - 4) = 7
  • Y-coordinate = 2 - (1 - 2) = 3
  • New coordinate C' = (7, 3)

Thus, the vertices of triangle A'B'C' after the transformation are A'(3, 3), B'(5, 1), C'(7, 3). This triangle should be drawn on the coordinate diagram.