Prove that the point (2,-2), (-2,1) and (5,2) are the vertices of the right angled triangle.
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Let the points be A(2, – 2), B(-2,1) and C(5,2)Applying distance formulaAB2 = (2 + 2)2\xa0+ (-2 – 1)2= 16 + 9AB2= 25 or, AB =5Similarly, BC2\xa0= (-2 – 5)2\xa0+ ( 1 – 2)2= 49 + 1 = 50or, BC2\xa0= 50 or, BC =\xa0{tex}\\sqrt{50}{/tex}Also,AC2\xa0= (2 – 5)2\xa0+ (-2 – 2)2= 9 + 16= 25or, AC2\xa0= 25 and AC = 5Clealry, AB2\xa0+ AC2\xa0= BC225 + 25 = 50Hence, the triangle is right – angled.Area of {tex}\\Delta{/tex}ABC = {tex}\\frac { 1 } { 2 } \\times \\text { Base } \\times \\text{Height}{/tex}{tex}= \\frac { 1 } { 2 } \\times 5 \\times 5 = \\frac { 25 } { 2 }{/tex}sq units.