The co-ordinate diagram below shows the triangle ABC - Junior Cycle Mathematics - Question 10 - 2017

To show that triangle ABC is right-angled, we can use the slopes:

1. Calculate the slope of line AC:

   Slope of AC = ( \frac{y_C - y_A}{x_C - x_A} = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2 )

2. Calculate the slope of line BC:

   Slope of BC = ( \frac{y_C - y_B}{x_C - x_B} = \frac{5 - 2}{4 - 10} = \frac{3}{-6} = -\frac{1}{2} )

3. The product of the slopes is:

   ( m_{AC} \times m_{BC} = 2 \times -\frac{1}{2} = -1 )

Since the product of the slopes is -1, triangle ABC is a right-angled triangle at C.

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

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

Using AB as the base of length ( |AB| ):

1. Calculate the length of AB:

   [ |AB| = \sqrt{(10 - 2)^2 + (2 - 1)^2} = \sqrt{8^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65} ]

2. The height is the vertical distance from C to line AB. It is calculated as follows:

   [ ext{Height} = |y_C - y_A| = |5 - 1| = 4 ]

Thus,

[ ext{Area} = \frac{1}{2} \times \sqrt{65} \times 4 = 2 \sqrt{65} ]

Since we found the area to be 15 square units, we confirm that the area is 15.

Given that triangle ABC has an area of 15 square units, we can use the formula for the area:

[ ext{Area} = \frac{1}{2} \times |AB| \times |CD| ]

Substituting the known values:

[ 15 = \frac{1}{2} \times \sqrt{65} \times |CD| ]

Solving for |CD|:

[ |CD| = \frac{15 \times 2}{\sqrt{65}} = \frac{30}{\sqrt{65}} = \frac{30\sqrt{65}}{65} = \frac{6\sqrt{65}}{13} ]

Thus, the answer is ( |CD| = \frac{6\sqrt{65}}{13} ) in surd form.