The points A (1, 2), B (3, 4) and C (4, 1) are the vertices of a triangle which is:

Correct option is C

​Given:

Vertices of triangle: $$A(1, 2), B(3, 4), C(4, 1)$$​

Formula used:

the lengths of sides using the distance formula:

​$$\text{Distance between } (x_1, y_1) \text{ and } (x_2, y_2): \,\\ \ \\ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$​​

Solution: 

Distance between A(1,2) and B(3, 4);

$$AB = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$ 

Distance between B(3, 4) and C(4, 1);

​$$BC = \sqrt{(4 - 3)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$$​​

Distance between C(4, 1) and A(1, 2)

​$$AC = \sqrt{(4 - 1)^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$$​​

Since $$BC = AC$$​ but $$AB \neq BC$$ ,

the triangle is isosceles.