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| 1. |
Find the coordinates of circumcentre |
| Answer» \xa0Let A(3, 0), B(-1, -6) and C(4, -1) be the given points.Let O(x, y) be the circumcentre of the triangle.OA = OB = OCOA2 = OB2(x - 3)2 + (y - 0)2 = (x + 1)2 + (y + 6 )2{tex}\\Rightarrow{/tex}\xa0x2 + 9 - 6x + y2 = x2 + 1 +2x + y2 + 36 + 12y{tex}\\Rightarrow{/tex}\xa0x2 - 6x + y2 - x2 - 2x - y2 - 12y = 1 + 36 - 9{tex}\\Rightarrow{/tex}\xa0-8x - 12y = 28{tex}\\Rightarrow{/tex}\xa0-2x - 3y = 7{tex}\\Rightarrow{/tex}\xa02x + 3y = -7 ........(i)Again,OB2 = OC2(x + 1)2 + (y + 6)2 = (x - 4)2 + (y + 1)2{tex}\\Rightarrow{/tex}\xa0x2 + 1 + 2x + y2 + 36 +12y = x2 +16 - 8x + y2 + 1 + 2y{tex}\\Rightarrow{/tex}\xa0x2 + 2x + y2 + 12y - x2 + 8x + y2 - 2y = 16 + 1 - 1 - 3610x + 10y = -20x + y = -2 ....... (ii)Solving (i) and (ii), we getx = 1, y = -3Hence circumcentre of the triangle is (1, -3)Circumradius\xa0{tex}= \\sqrt { ( 1 + 1 ) ^ { 2 } + ( - 3 + 6 ) ^ { 2 } }{/tex}{tex}= \\sqrt { ( 2 ) ^ { 2 } + ( 3 ) ^ { 2 } }{/tex}{tex}= \\sqrt { 4 + 9 }{/tex}{tex}= \\sqrt { 13 }{/tex}\xa0units. | |