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Boa noite Iuri!
Solução!
Geometria analítica formula da distância entre dois pontos.
![\sqrt{( 2 \sqrt{3} - 4 \sqrt{3} )^{2}+(y-1)^{2}}=4\\\\\\
\sqrt{( 2 - 4 )\sqrt{3} )^{2}+(y-1)^{2}}=4\\\\\\
\sqrt{( -2 \sqrt{3} )^{2}+(y-1)^{2}}=4\\\\\\
\sqrt{( 4.3 )^{2}+(y^{2}-2y+1 )}=(4)^{2} \\\\\\
12 +y^{2}-2y+1 =16\\\\\\\
y^{2}-2y+1-16+12=0\\\\\\
y^{2}-2y-3=0](https://tex.z-dn.net/?f=+%5Csqrt%7B%28+2+%5Csqrt%7B3%7D+-+4+%5Csqrt%7B3%7D+%29%5E%7B2%7D%2B%28y-1%29%5E%7B2%7D%7D%3D4%5C%5C%5C%5C%5C%5C%0A++%5Csqrt%7B%28+2++-+4+%29%5Csqrt%7B3%7D+%29%5E%7B2%7D%2B%28y-1%29%5E%7B2%7D%7D%3D4%5C%5C%5C%5C%5C%5C+%0A+%5Csqrt%7B%28+-2+%5Csqrt%7B3%7D+%29%5E%7B2%7D%2B%28y-1%29%5E%7B2%7D%7D%3D4%5C%5C%5C%5C%5C%5C+%0A+%5Csqrt%7B%28+4.3+%29%5E%7B2%7D%2B%28y%5E%7B2%7D-2y%2B1+%29%7D%3D%284%29%5E%7B2%7D+%5C%5C%5C%5C%5C%5C%0A12+%2By%5E%7B2%7D-2y%2B1+%3D16%5C%5C%5C%5C%5C%5C%5C%0Ay%5E%7B2%7D-2y%2B1-16%2B12%3D0%5C%5C%5C%5C%5C%5C%0Ay%5E%7B2%7D-2y-3%3D0 "\sqrt{( 2 \sqrt{3} - 4 \sqrt{3} )^{2}+(y-1)^{2}}=4\\\\\\
\sqrt{( 2 - 4 )\sqrt{3} )^{2}+(y-1)^{2}}=4\\\\\\
\sqrt{( -2 \sqrt{3} )^{2}+(y-1)^{2}}=4\\\\\\
\sqrt{( 4.3 )^{2}+(y^{2}-2y+1 )}=(4)^{2} \\\\\\
12 +y^{2}-2y+1 =16\\\\\\\
y^{2}-2y+1-16+12=0\\\\\\
y^{2}-2y-3=0")
![a=1\\\\
b=-2\\\\
c=-3\\\\\
y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a} \\\\\\
y=\dfrac{-(-2)\pm \sqrt{(-2)^{2}-4.1.(-3) } }{2.1} \\\\\\
y=\dfrac{2\pm \sqrt{4+12} }{22} \\\\\\
y=\dfrac{2\pm \sqrt{16 } }{2} \\\\\\
y=\dfrac{2\pm 4 }{2} \\\\\\
y_{1}= \dfrac{2+4}{2} = \dfrac{6}{2}=3 \\\\\\\
y_{2}= \dfrac{2-4}{2} = \frac{-2}{2} =-1](https://tex.z-dn.net/?f=%0Aa%3D1%5C%5C%5C%5C%0Ab%3D-2%5C%5C%5C%5C%0Ac%3D-3%5C%5C%5C%5C%5C%0Ay%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E%7B2%7D-4.a.c+%7D+%7D%7B2.a%7D+%5C%5C%5C%5C%5C%5C%0Ay%3D%5Cdfrac%7B-%28-2%29%5Cpm+%5Csqrt%7B%28-2%29%5E%7B2%7D-4.1.%28-3%29+%7D+%7D%7B2.1%7D+%5C%5C%5C%5C%5C%5C%0Ay%3D%5Cdfrac%7B2%5Cpm+%5Csqrt%7B4%2B12%7D+%7D%7B22%7D+%5C%5C%5C%5C%5C%5C%0A%0Ay%3D%5Cdfrac%7B2%5Cpm+%5Csqrt%7B16+%7D+%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%0A%0Ay%3D%5Cdfrac%7B2%5Cpm+4+%7D%7B2%7D+%5C%5C%5C%5C%5C%5C%0A%0Ay_%7B1%7D%3D+%5Cdfrac%7B2%2B4%7D%7B2%7D+%3D+%5Cdfrac%7B6%7D%7B2%7D%3D3+%5C%5C%5C%5C%5C%5C%5C%0Ay_%7B2%7D%3D+%5Cdfrac%7B2-4%7D%7B2%7D+%3D+%5Cfrac%7B-2%7D%7B2%7D+%3D-1++ "a=1\\\\
b=-2\\\\
c=-3\\\\\
y= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a} \\\\\\
y=\dfrac{-(-2)\pm \sqrt{(-2)^{2}-4.1.(-3) } }{2.1} \\\\\\
y=\dfrac{2\pm \sqrt{4+12} }{22} \\\\\\
y=\dfrac{2\pm \sqrt{16 } }{2} \\\\\\
y=\dfrac{2\pm 4 }{2} \\\\\\
y_{1}= \dfrac{2+4}{2} = \dfrac{6}{2}=3 \\\\\\\
y_{2}= \dfrac{2-4}{2} = \frac{-2}{2} =-1")
Os dois valores de y são verdadeiro para a distância.
Boa noite!
Bons estudos!
Solução!
Geometria analítica formula da distância entre dois pontos.
Os dois valores de y são verdadeiro para a distância.
Boa noite!
Bons estudos!
iuridaniel:
Obg ♥
Dê nada!
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