Question

Prove that A(0, 7), B(4, 3), C(6, 5) form the vertices of a right-angled triangle.

Answer 1

Given:

Points A(0, 7), B(4, 3), and C(6, 5).

Triangle ABC is a right-angled triangle.

Step 1: Find the lengths of the sides AB, BC, and AC using the distance formula

The distance formula between two points (x₁, y₁) and (x₂, y₂) is:

`"Distance" = sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`

Length (AB):

\[\sqrt{(4 - 0)^2 + (3 - 7)^2}\]

= \[\sqrt{4^2 + (-4)^2}\]

= \[\sqrt{16 + 16}\] 

= \[\sqrt{32}\] 

= \[4\sqrt{2}\]

Length (BC):

\[\sqrt{(6 - 4)^2 + (5 - 3)^2}\]

= \[\sqrt{2^2 + 2^2}\]

= \[\sqrt{4 + 4}\]

= \[\sqrt{8}\]

= \[2\sqrt{2}\]

Length (AC):

\[\sqrt{(6 - 0)^2 + (5 - 7)^2}\]

= \[\sqrt{6^2 + (-2)^2}\]

= \[\sqrt{36 + 4}\] 

= \[\sqrt{40}\] 

= \[2\sqrt{10}\]

Step 2: Check the Pythagorean theorem condition

For a right-angled triangle, one side squared equals the sum of the squares of the other two sides.

\[AB^2 = (4\sqrt{2})^2 = 16 \times 2 = 32\]

\[BC^2 = (2\sqrt{2})^2 = 4 \times 2 = 8\]

\[AC^2 = (2\sqrt{10})^2 = 4 \times 10 = 40\]

\[AB^2 + BC^2 = AC^2\]

\[32 + 8 = 40 \quad \Rightarrow \quad 40 = 40\]

Triangle ABC with vertices A(0, 7), B(4, 3), and C(6, 5) is right-angled at B because (AB² + BC² = AC²) satisfies the Pythagorean theorem.