Prove that vectors BA and BC are perpendicular if A (0; 1), B (2; -1), C (4; 1)

Using the formula for the distance between two points on the coordinate plane, we find the lengths of the vectors BA, BC and AC:

| BA | = √ ((2 – 0) ^ 2 + (-1 – 1) ^ 2) = √ (2 ^ 2 + (-2) ^ 2) = √ (2 ^ 2 + 2 ^ 2) = √ (4 + 4) = √8;

| BC | = √ ((2 – 4) ^ 2 + (-1 – 1) ^ 2) = √ ((- 2) ^ 2 + (-2) ^ 2) = √ (2 ^ 2 + 2 ^ 2) = √ (4 + 4) = √8;

| AC | = √ ((4 – 0) ^ 2 + (1 – 1) ^ 2) = √ (4 ^ 2 + 0 ^ 2) = √ (16 + 0) = √16 = 4.

Check if the relation | AC | ^ 2 = | BA | ^ 2 + | BC | ^ 2 is satisfied:

4 ^ 2 = (√8) ^ 2 + (√8) ^ 2:

16 = 8 + 8;

16 = 16.

Since the relation | AC | ^ 2 = | BA | ^ 2 + | BC | ^ 2 is satisfied, the triangle ABC is right-angled and the angle ABC is right.

Therefore, vectors BA and BC are perpendicular.