Question

O ponto B tem ordenada nula e dista 5 de A, que possui ambas as coordenadas iguais a 4. Ache as abscissa de B.

Answer 1

$B\left(x_{_{B}},\,0 \right ) \Rightarrow y_{_{B}}=0\\ \\ A\left(4,\,4 \right ) \Rightarrow \left\{ \begin{array}{c} x_{_{A}}=4\\ y_{_{A}}=4 \end{array} \right.$


A distância entre os pontos $A$ e $B$ é dada por

$d_{_{AB}}=\sqrt{\left(x_{_{B}}-x_{_{A}} \right )^{2}+\left(y_{_{B}}-y_{_{A}} \right )^{2}}\\ \\ \\ 5=\sqrt{\left(x_{_{B}}-4 \right )^{2}+\left(0-4 \right )^{2}}\\ \\ 5=\sqrt{\left(x_{_{B}}-4 \right )^{2}+\left(-4 \right )^{2}}\\ \\ 5=\sqrt{\left(x_{_{B}}-4 \right )^{2}+16}\\ \\ \left(5\right)^{2}=\left(\sqrt{\left(x_{_{B}}-4 \right )^{2}+16} \right )^{2}\\ \\ 25=\left(x_{_{B}}-4 \right )^{2}+16\\ \\ \left(x_{_{B}}-4 \right )^{2}=25-16\\ \\ \left(x_{_{B}}-4 \right )^{2}=9\\ \\ x_{_{B}}-4=\pm \sqrt{9}\\ \\ x_{_{B}}-4=\pm 3\\ \\ x_{_{B}}=\pm 3 -4\\ \\ \begin{array}{rcl} x_{_{B}}=3-4&\text{ ou }&x_{_{B}}=-3-4\\ \end{array}\\ \\ \boxed{\begin{array}{rcl} x_{_{B}}=-1&\text{ ou }&x_{_{B}}=-7\\ \end{array}}$