Find the point on the x-axis which is equidistant from (2, –5)

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 Multiple Choice QuestionsShort Answer Type

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121.

Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9)


Let the required point be ‘P’ which is on the x-axis, so its ordinate = 0 and abscissa (say) = x
Therefore, co-ordinates of the point P are (x, 0).
Let the given points be A(2, -5) and B(—2, 9).
It is given that:

space
space space space space space space AP space equals space BP
rightwards double arrow space space space space square root of left parenthesis 2 minus straight x right parenthesis squared plus left parenthesis negative 5 minus 0 right parenthesis squared end root
space space space space equals space space space square root of left curly bracket straight x minus left parenthesis negative 2 right parenthesis right curly bracket squared plus left parenthesis 0 minus 9 right parenthesis squared end root
rightwards double arrow square root of left parenthesis 2 minus straight x right parenthesis squared plus left parenthesis negative 5 right parenthesis squared end root equals square root of open parentheses straight x minus 2 close parentheses squared plus left parenthesis negative 9 right parenthesis squared end root
rightwards double arrow space square root of left parenthesis 2 right parenthesis squared left parenthesis straight x right parenthesis squared minus 2 left parenthesis 2 right parenthesis left parenthesis straight x right parenthesis plus 25 end root
equals space square root of left parenthesis straight x right parenthesis squared plus left parenthesis 2 right parenthesis squared plus 2 left parenthesis straight x right parenthesis left parenthesis 2 right parenthesis plus 81 end root
rightwards double arrow space square root of 4 plus straight x squared minus 4 straight x plus 25 end root equals square root of straight x squared plus 4 plus 4 straight x plus 81 end root
rightwards double arrow space square root of straight x squared minus 4 straight x plus 29 end root equals square root of straight x squared plus 4 straight x plus 85 end root

x2 - 4x + 29 = x2 + 4x + 85
⇒    -4x - 4x = 85 - 29
⇒    -8x = 56⇒    x = -7
Therefore, the required point is (-7, 0).

 
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Fig. 7.12

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