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| 1. |
Find a point on the y-axis which is equidistant from the points A(6,5) and B(-4,3). |
| Answer» We have to find a point\xa0on the y-axis which is equidistant from the points A(6,\xa05) and B (- 4, 3).We know that a point on y-axis is of the form (0, y). So, let the required point be P (0, y).Then,PA = PB{tex}\\Rightarrow \\sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \\sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}\xa036 + (y - 5)2 = 16 + (y - 3)2{tex}\\Rightarrow{/tex}\xa036 + y2\xa0- 10y + 25 = 16 + y2 - 6y + 9{tex}\\Rightarrow{/tex}\xa04y = 36{tex}\\Rightarrow{/tex}\xa0y = 9So, the required point is (0, 9). | |