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 calculate the distances between the points A, B, and C, followed by applying the Pythagorean theorem.

1. Calculate the distances:

   • For AB:

   $AB = \sqrt{(8-15)^2 + (17-10)^2} = \sqrt{(-7)^2 + (7)^2} = \sqrt{49 + 49} = \sqrt{98}$

   • For BC:

   BC = \sqrt{(15 - (-2))^2 + (10 - (-7))^2 = \sqrt{(15 + 2)^2 + (10 + 7)^2} = \sqrt{17^2 + 17^2} = \sqrt{289 + 289} = \sqrt{578}

   • For AC:

   $AC = \sqrt{(8 - (-2))^2 + (17 - (-7))^2} = \sqrt{(8 + 2)^2 + (17 + 7)^2} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26$

2. Now apply the Pythagorean theorem:

   • Check if:

   $AB^2 + BC^2 = AC^2$

   • Substitute the distances:

   $98 + 578 = 676$

   • Therefore, since 676 = 676, angle ABC is a right angle.