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tulipah
@tulipah
July 2022
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На прямой 3x-y+4=0 найти точку , равноудаленную от точек A(3; 3) и B(7; 5). Ответ должен быть : (x=2, y=10).
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shavilov999110
Y=3x+4
(x-3)^2+(y-3)^2=(x-7)^2+(y-5)^2
x^2-6x+9+y-6y+9=x^2-14x+49+y^2-10y+25
-6x-6y+18=-14x-10y+74
8x=-4y+56
x=7-1/2y
y=3(7-1/2y)+4
y=21-3/2y+4
2,5y=25
y=10
y=3x+4
3x+4=10
3x=6 x=6/3 x=2
x=2 y=10
1 votes
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Answers & Comments
(x-3)^2+(y-3)^2=(x-7)^2+(y-5)^2
x^2-6x+9+y-6y+9=x^2-14x+49+y^2-10y+25
-6x-6y+18=-14x-10y+74
8x=-4y+56
x=7-1/2y
y=3(7-1/2y)+4
y=21-3/2y+4
2,5y=25
y=10
y=3x+4
3x+4=10
3x=6 x=6/3 x=2
x=2 y=10