Question

3. Determina las coordenadas de un punto que equi-
diste de los puntos A (1, 3), B (3, 1) y C (-3,-1).

Answer 1

Respuesta:

Explicación

Sea p el puno a encontrar

tomamos la distancia de PA y la igualamos con l distancia de PB

$\\SEA\\PA=PB=PC\\\\\\1) PA=PB\\\sqrt{(x-1)^{2}+(y-3)^{2} } = \sqrt{(x-3)^{2}+(y-1)^{2}} \\\\(x-1)^{2}+(y-3)^{2} =(x-3)^{2}+(y-1)^{2}\\x^{2} -2x+1+y^{2} -6y+9=x^{2} -6x+9+y^{2} -2y+1\\ -2x+1-6y+9= -6x+9 -2y+1\\-2x+6x-6y+2y=1+9-9-1\\4x-4y=0\\x-y=0\\x=y..................................... (1)\\\\2) PA=PC\\\sqrt{(x-1)^{2}+(y-3)^{2} } = \sqrt{(x-(-3))^{2}+(y-(-1))^{2}} \\(x-1)^{2}+(y-3)^{2} =(x+3)^{2}+(y+1)^{2}\\x^{2} -2x+1+y^{2} -6y+9=x^{2} +6x+9+y^{2} +2y+1\\-2x+1+y^{2} -6y+9=x^{2} +6x+9+y^{2} +2y+1$$-2x-6x-6y-2y=1+9-9-1\\-8x-8y=0\\-x-y=0\\-x=y................................(2)$

con las ecuaciones (1) y (2)

tenemos el sistema de ecuaciones $\left \{ {{x=y} \atop {-x=y}} \right. \\y=x=-x\\ \Rightarrow\left \{ {{x=0} \atop {y=0} \right. \\\\\therefore P(0,0)\\comprobacion \\PA=\sqrt{(0-1)^2\\+(0-3)^2} = \sqrt{1+9} =\sqrt{10} \\PB=\sqrt{(0-3)^2\\+(0-1)^2} = \sqrt{9+1} =\sqrt{10} \\PC=\sqrt{(0-(-3))^2\\+(0-(-1))^2} =\sqrt{(0+3)^2\\+(0+1)^2}= \sqrt{9+1} =\sqrt{10} \\\\\therefore PA=PB=PC$