Question

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

Answer 1

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

$\to$ 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 )