Answer :

Formula used:


(10, 0), (18, 0) and (10, 15)


Let the points be A (10, 15), B (10, 0) and C (18, 0)


Distance of AB


AB = √ ((10 – 10))2 + (0 – 15)2)


AB = √ ((0)2 + (–15)2)


AB = √ (0 + 225)


AB = √225


Distance of BC


B C= √ ((18 – 10)2 + (0 – 0)2)


BC = √ ((8)2 + (0)2)


BC = √ (64 + 0)


BC = √ 64


Distance of AC


AC = √ ((18 – 10)2 + (0 – 15))2)


AC = √ ((8)2 + (–15)2)


AC = √ (64 + 225)


AC = √289


i.e. AB2 + BC2


= (√225)2 + (√64)2


= 225 + 64


= 289 = (AC)2


Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.


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