Question

determine as coordenadas dos pontos do eixo das abscissas que distam 5 do ponto P (6, -3) Q(x,0)

Answer 1

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)}^2 \\ 5 = \sqrt{(x-6)^2+(0-(-3))^2} \\ 5 = \sqrt{x^2-12x+36+(3)^2} \\ 5 = \sqrt{x^2-12x+36+9} \\ 5 = \sqrt{x^2-12x+45} \\ \{5 = \sqrt{x^2-12x+45} \}^2\\ 25=x^2-12x+45 \\ x^2-12x+45-25=0\\ x^2-12x+20=0\\ \\ \Delta=b^2-4ac\\ \Delta=(-12)^2-4.1.20\\ \Delta=144-80\\ \Delta=64\\ \\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \\ \\ x= \dfrac{12\pm \sqrt{64} }{2} \\ \\ x= \dfrac{12\pm8}{2} \\ \\ x=6\pm4 \\ \\ x'=6+4=10\\ x"=6-4=2$

Ponto Q = {10,0} ou Q={2,0}

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