Answer: The calculations are done below.
Step-by-step explanation:
(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,
![AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%282-3%29%5E2%2B%280-2%29%5E2%7D%3D%5Csqrt%7B5%7D%2C%5C%5C%5C%5CBC%3D%5Csqrt%7B%283-5%29%5E2%2B%282-1%29%5E2%7D%3D%5Csqrt%7B5%7D%2C%5C%5C%5C%5CCA%3D%5Csqrt%7B%285-2%29%5E2%2B%281-0%29%5E2%7D%3D%5Csqrt%7B10%7D.)
Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.
(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,
![AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%284-6%29%5E2%2B%282-2%29%5E2%7D%3D%5Csqrt%7B4%7D%3D2%2C%5C%5C%5C%5CBC%3D%5Csqrt%7B%286-5%29%5E2%2B%282-3.73%29%5E2%7D%3D%5Csqrt%7B14.3729%7D%2C%5C%5C%5C%5CCA%3D%5Csqrt%7B%285-4%29%5E2%2B%283.73-2%29%5E2%7D%3D%5Csqrt%7B14.3729%7D.)
Since, BC = CA, so the triangle ABC will be an isosceles triangle.
(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,
![AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-5%2B4%29%5E2%2B%282-4%29%5E2%7D%3D%5Csqrt%7B5%7D%2C%5C%5C%5C%5CBC%3D%5Csqrt%7B%28-4%2B2%29%5E2%2B%284-2%29%5E2%7D%3D%5Csqrt%7B8%7D%2C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-2%2B5%29%5E2%2B%282-2%29%5E2%7D%3D%5Csqrt%7B9%7D.)
Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.
(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,
![AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-3%2B3%29%5E2%2B%281-4%29%5E2%7D%3D%5Csqrt%7B9%7D%3D3%2C%5C%5C%5C%5CBC%3D%5Csqrt%7B%28-3%2B1%29%5E2%2B%284-1%29%5E2%7D%3D%5Csqrt%7B13%7D%2C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-1%2B3%29%5E2%2B%281-1%29%5E2%7D%3D%5Csqrt%204.)
Since AB² + CA² = BC², so this will be a right angled triangle.
(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,
![AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-4%2B2%29%5E2%2B%282-4%29%5E2%7D%3D%5Csqrt%7B8%7D%2C%5C%5C%5C%5CBC%3D%5Csqrt%7B%28-2%2B1%29%5E2%2B%284-4%29%5E2%7D%3D%5Csqrt%7B1%7D%3D1%2C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-1%2B4%29%5E2%2B%284-2%29%5E2%7D%3D%5Csqrt%7B13%7D.)
Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.
Thus, the match is done.