Math, asked by harsh1692, 10 months ago

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

Answers

Answered by Anonymous
2

 \large \bold{ \underline{ \underline{ \: Answer : \:  \:  \: }}}

 \to The Required point is P( - 7 , 0 )

\large \bold{ \underline{ \underline{ \:  Explaination  : \:  \:  \: }}}

Let , the required point be P( x , 0 )

Given ,

\bold{X_{1}  = 2 } \\ </p><p>\bold {X_{2} = -2 }    \\  \bold{ Y_{1} = -5  } \\ </p><p>\bold{ Y_{2} = 9 }

  \bold{Distance \:  Formula} =  \sqrt{( X_{2} - {X_{1}   </p><p>)}^{2} +  {(Y_{2} - {Y_{1}   </p><p>)}}^{2}   \:  \: }

According to the question ,

AP = PB

 \to \sqrt{ {(x - 2)}^{2}  +  {(0 + 5)}^{2} }   =  \sqrt{ {(x + 2)}^{2} +  {(0 - 9)}^{2}  }  \\  \\  \to {x}^{2}  + 4 - 4x + 25 =  {x}^{2}  + 4 + 4x + 81 \\  \\   \to - 4x + 25 = 4x + 81 \\  \\   \to- 8x = 56 \\  \\  \to x =   - 7

Hence , P ( - 7 , 0 )

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