find the point x-axis, when is equidistant from (2, 3) & (-4, -5), (4, 5)
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Answer:
AC=BC (Equidistance)
AC
2
=BC
2
(x−5)
2
+(0−4)
2
=(x+2)
2
+(0−3)
2
x
2
−10x+25+16=x
2
+4+4x+9
−14x+41−13=0
−14x+28=0
=14x=−28
x=28/14
x=2
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