Find them coordinates of them point equidistant from three given points A (5,1),B(-3,-7)and (7,-1)
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We haveA {tex}\\rightarrow{/tex} (5, 1)B {tex}\\rightarrow{/tex} (-3, -7)C {tex}\\rightarrow{/tex} (7, -1)Let the point P(x, y) be equidistant from the three given points A, B, and C.Then PA = PB = PC{tex}\\Rightarrow{/tex} PA2 = PB2 = PC2First two givePA2 = PB2{tex}\\Rightarrow{/tex} (x – 5)2 + (y – 1)2 = (x + 3)2 + (y + 7)2{tex}\\Rightarrow{/tex} x2 – 10x + 25 + y2 – 2y + 1 = x2 + 6x + 9 + y2 + 14y + 49{tex}\\Rightarrow{/tex} 16x + 16y + 32 = 0{tex}\\Rightarrow{/tex} x + y + 2 = 0 ….Dividing throughout by 16 ….(1)Last two givePB2 = PC2{tex}\\Rightarrow{/tex} (x + 3)2 + (y + 7)2 = (x – 7)2 + (y + 1)2{tex}\\Rightarrow{/tex} x2 + 6x + 9 + y2 + 14y + 49 = x2 – 14x + 49 + y2 + 2y + 1{tex}\\Rightarrow{/tex} 20x + 12y + 8 = 0{tex}\\Rightarrow{/tex} 5x + 3y + 2 = 0 ….Dividing throughout by 4 ….(2)Multiplying equation (1) by 3, we get3x + 3y + 6 = 0 ….(3)Subtracting equation (3) from equation (2), we get2x – 4 = 0{tex}\\Rightarrow{/tex} 2x = 4{tex}\\Rightarrow x = \\frac{4}{2} = 2{/tex}Subtracting x = 2 in equation (1), we get2 + y + 2 = 0{tex}\\Rightarrow{/tex} y + 4 = 0{tex}\\Rightarrow{/tex} y = -4Hence, the required points is (2, -4).