Determine the type of triangle ABC if A (2; 1) B (8; 7) C (9; 0)

Using the formula for the distance between two points on the coordinate plane, we find the lengths of the sides of this triangle ABC:

| AB | = √ ((8 – 2) ^ 2 + (7 – 1) ^ 2) = √ (6 ^ 2 + 6 ^ 2) = √ (2 * 6 ^ 2) = 6√2;

| BC | = √ ((8 – 9) ^ 2 + (7 – 0) ^ 2) = √ (1 ^ 2 + 7 ^ 2) = √ (1 + 49) = √50:

| AC | = √ ((9 – 2) ^ 2 + (0 – 1) ^ 2) = √ (7 ^ 2 + 1 ^ 2) = √ (49 + 1) = √50.

Since the lengths of the sides BC and AC are equal, this triangle is isosceles.

Applying the cosine theorem, we find the angle cos (C):

(√50) ^ 2 + (√50) ^ 2 – 2 * √50 * √50 * cos (C) = (6√2) ^ 2:

50 + 50 – 100 * cos (C) = 36 * 2;

100 – 100 * cos (C) = 72;

100 * cos (C) = 100 – 72;

100 * cos (C) = 28;

cos (C) = 28/100 = 0.28.

Since cos (C)> 0, this triangle is acute-angled.

Answer: This triangle is isosceles and acute-angled.