find the point on the x-axis which is equidistant from point (5, 4) and (-2,3)
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αc=вc (єquídíѕtαncє)
αc²=вc²
(х−5)²+(0−4)²
=(х+2)²+(0−3)²
х²−10х+25+16=х²+4+4х+9
−14х+41−13=0
−14х+28=0
=14х=−28
х=28/14
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