the points
on the x-aris which is equidistor
from (5,-5) and (-2,9)​

1 thought on “the points<br />on the x-aris which is equidistor<br />from (5,-5) and (-2,9)​”

1. Answer:

   (-2.5 , 0)

   Step-by-step explanation:

   distances are equal from both the points so

   [tex]\sqrt{(x1-x)^{2}+(y1-y)^2 } = \sqrt{(x2-x)^{2}+(y2-y)^2 }[/tex]

   [tex]\sqrt{(5-x)^{2}+(-5-y)^2 } = \sqrt{(-2-x)^{2}+(9-y)^2 }[/tex]

   as the point is in the x-axis then y=0, so

   [tex]\sqrt{(5-x)^{2}+(-5)^2 } = \sqrt{(-2-x)^{2}+(9)^2 }[/tex]

   [tex]\ (5-x)^{2}+(-5)^2 } = \ (-2-x)^{2}+(9)^2 }[/tex]

   after solving this we get

   x = -2.5

   so the final coordinate is (-2.5, 0).

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