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| 1. |
Find the coordinates of the point equidistant from 3given points A(5,1) , B(-3,-1) & C(7,-1) |
| Answer» 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). | |