Classifying Triangle Vertices

To determine the type of triangle formed by points A (1, 2), B (3, 4), and C (4, 1), we need to calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Calculating Side Lengths

Calculate the length of each side:

• Side AB: $d_{AB} = \sqrt{(3 - 1)^2 + (4 - 2)^2}$ $d_{AB} = \sqrt{2^2 + 2^2}$ $d_{AB} = \sqrt{4 + 4}$ $d_{AB} = \sqrt{8}$
• Side BC: $d_{BC} = \sqrt{(4 - 3)^2 + (1 - 4)^2}$ $d_{BC} = \sqrt{1^2 + (-3)^2}$ $d_{BC} = \sqrt{1 + 9}$ $d_{BC} = \sqrt{10}$
• Side AC: $d_{AC} = \sqrt{(4 - 1)^2 + (1 - 2)^2}$ $d_{AC} = \sqrt{3^2 + (-1)^2}$ $d_{AC} = \sqrt{9 + 1}$ $d_{AC} = \sqrt{10}$

Determining Triangle Type

Compare the side lengths:

• $AB = \sqrt{8}$
• $BC = \sqrt{10}$
• $AC = \sqrt{10}$

Since two sides, BC and AC, have equal lengths ($\sqrt{10}$), the triangle is isosceles.

We can also check if it's a right-angled triangle using the Pythagorean theorem ($a^2 + b^2 = c^2$). The squares of the side lengths are 8, 10, and 10.

• Does $8 + 10 = 10$? No.
• Does $10 + 10 = 8$? No.

The triangle does not satisfy the Pythagorean theorem, so it is not right-angled. As it has two equal sides and is not equilateral or right-angled, it is classified as isosceles.