Question

O ponto P(X,-3) É equidistante de A(4,1) e B(-2,-1). Encontre o valor de x

Answer 1

Se são equidistantes, a distancia AP e a distancia BP deve ser igual.

A distancia entre dois pontos é dada por:

$\boxed{Distancia_{A,B}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}}$

Temos então:

$Distancia_{A,P}~=~Distancia_{B,P}\\\\\\\sqrt{(4-x)^2+(1-(-3))^2}~=~\sqrt{(-2-x)^2+(-1-(-3))^2}\\\\\\(4-x)^2+(1-(-3))^2~=~(-2-x)^2+(-1-(-3))^2\\\\\\(x^2-8x+16)+(16)~=~(x^2+4x+4)+(4)\\\\\\x^2-x^2-8x-4x~=~4+4-16-16\\\\\\-12x~=~-24\\\\\\x~=~\frac{-24}{-12}\\\\\\\boxed{x~=~2}$