Question

find the point on tge x axis which is equidistant from (2, -5) , and (-2, 9)?

Answer 1

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Answer 2

$\textbf{Let the point be ( x , 0) , } \\ \\ \textbf{By Distance formula, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: distance \: formula } \rightarrow\: \: distance \: between \: two \: points \: = \sqrt{(x_{1} - x_{2} )^{2} + ( y_{1} - y_{2} ) ^{2}} \\ \\ \\ \textbf{ Distance between ( 2, - 5 ) and ( x , y ) = Distance between ( x , y ) and ( - 2 , 9 ) } \\ \\ \\ Hence, \\ \\ \sqrt{ {(x - 2)}^{2} + {(0+ 5)}^{2} } = \sqrt{ {( - 2 - x)}^{2} + {(9 - 0)}^{2} } \\ \\ Square \: on \: both \: sides, \\ \\ \\ \\ = > {(x - 2)}^{2} + {5}^{2} = {(2 + x)}^{2} + {9}^{2} \\ \\ = > {x}^{2} + 4 - 4x + 25 = 4 + {x}^{2} + 4x + 81 \\ \\ = > 25 - 81 = 4x +4x \\ \\ = > - 56 = 8x \\ \\ = > - \frac{56}{8} = x \\ \\ = > - 7 = x$



Point = ( 7 , 0 )