Question

Considere os pontos A(-1;5), B(3;7) e P(x;0). Sabendo que a distância entre P e A(dPA) é igual à distância entre P e B (dPB), o valor de x é:

Answer 1

$d(PA)=d(PB)\\\\\\\sqrt{(xp-xa)^2+(yp-ya)^2}=\sqrt{(xp-xb)^2+(yp-yb)^2}\\(x-[-1])^2+(0-5)^2=(x-3)^2+(0-7)^2\\(x+1)^2+25=(x-3)^2+49\\x^2+2x+1+25=x^2-6x+9+49\\2x+6x+26=58\\8x=58-26=32\\\\\boxed{x=\frac{32}{8}=4}\\\\\\P(4,0)$