Question

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

Answer 1

$\mathtt{ \huge{ \fbox{SOLUTION : }}}$

Let , Point B be the point whose coordinate is (x,0) which is equidistant from A(2,-5) and C(-2,9)

We know that , the distance between two points is given by

$\large \fbox{ \sf D = \sqrt{ { ( x_{2} - x_{1} )}^{2} + {( y_{2} - y_{1} )}^ {2} } \: \: }$

According to the question ,

$\sf \hookrightarrow \sqrt{ {(x - 2)}^{2} + {(0 - ( - 5))}^{2} } = \sqrt{ { (-2 - x)}^{2} + {(9 - 0)}^ {2} } \\ \\ \sf \hookrightarrow \sqrt{{x}^{2} + {( - 2)}^{2} - 2x (2 )+ 25} = \sqrt{{( - 2)}^{2} + {(x)}^{2} - 2 ( - 2)x + 81} \\ \\ \sf \hookrightarrow \sqrt{ {(x)}^{2} + 4 - 4x + 25 } = \sqrt{ {(x)}^{2} + 4 + 4x + 85} \\ \\ \sf \hookrightarrow \sqrt{ {(x)}^{2} - 4x + 29} = \sqrt{ {(x)}^{2} + 4x + 85 }$

On Squaring both sides , we obtain

$\sf \hookrightarrow {(x)}^{2} - 4x + 29= {(x)}^{2} + 4x + 85 \\ \\ \sf \hookrightarrow 8x = - 56 \\ \\\sf \hookrightarrow x = - 7$

Hence , B(-7,0) is the points on the x axis which is equidistant from (2,-5) and (-2,9)