Three points A, B and C have coordinates A (8, 17), B (15, 10) and C (–2, –7) - AQA - A-Level Maths Pure - Question 7 - 2018 - Paper 1

To show that angle ABC is a right angle, we will determine the slopes of lines AB and BC, and check if their product is equal to -1.

1. Calculate the coordinates:

   • A (8, 17)
   • B (15, 10)
   • C (–2, –7)

2. Determine the lengths:
   Using the distance formula,
   $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

   • Distance AB:
     $AB = \sqrt{(15 - 8)^2 + (10 - 17)^2} = \sqrt{(7)^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98}$
   • Distance BC:
     $BC = \sqrt{(–2 - 15)^2 + (–7 - 10)^2} = \sqrt{(-17)^2 + (-17)^2} = \sqrt{289 + 289} = \sqrt{578}$
   • Distance AC:
     $AC = \sqrt{(–2 - 8)^2 + (–7 - 17)^2} = \sqrt{(-10)^2 + (-24)^2} = \sqrt{100 + 576} = \sqrt{676} = 26$

3. Calculate the slopes:

   • Slope of AB:
     $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{10 - 17}{15 - 8} = \frac{-7}{7} = -1$
   • Slope of BC:
     $m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 10}{-2 - 15} = \frac{-17}{-17} = 1$

4. Check the product of slopes:

   • The product of slopes is:
     $m_{AB} * m_{BC} = (-1)(1) = -1$

Since the product of the slopes is -1, we conclude that angle ABC is a right angle.