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sasuke97
@sasuke97
July 2022
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уравнение геометрического места точек на плоскости OXY, равноудаленных от точек А(5:4) и В (7:-2) имеет вид
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(x-5)^2+(y-4)^2=(x-7)^2+(y+2)^2
x^2-10x+25+y^2-8y+16=x^2-14x+49+y^2+4y+4
-10x+25-8y+16=-14x+49+4y+4
4x-12y+41=53
4x-12y=12
x-3y=3
y=(x/3)-1
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Verified answer
(x-5)^2+(y-4)^2=(x-7)^2+(y+2)^2x^2-10x+25+y^2-8y+16=x^2-14x+49+y^2+4y+4
-10x+25-8y+16=-14x+49+4y+4
4x-12y+41=53
4x-12y=12
x-3y=3
y=(x/3)-1