Question

O ponto medio de um segmento PQ e M (5,4). Quais são as coordenadas de Q se:
A) P(-3,2)
B) P (0,0)

Answer 1

Se $M\left(x_{_{M}},\,y_{_{M}} \right )$ é o ponto médio do segmento formado pelos pontos $P\left(x_{_{P}},\,y_{_{P}} \right )$ e $Q\left(x_{_{Q}},\,y_{_{Q}} \right )$, então

$x_{_{M}}=\dfrac{x_{_{P}}+x_{_{Q}}}{2}\\ \\ y_{_{M}}=\dfrac{y_{_{P}}+y_{_{Q}}}{2}$


Para as questões a seguir, temos que o ponto médio é $M\left(5,\,4 \right )$:


a) $P\left(-3,\,2 \right )$

$x_{_{M}}=\dfrac{x_{_{P}}+x_{_{Q}}}{2}\\ \\ 5=\dfrac{-3+x_{_{Q}}}{2}\\ \\ -3+x_{_{Q}}=2\cdot 5\\ \\ -3+x_{_{Q}}=10\\ \\ x_{_{Q}}=10+3\\ \\ \boxed{x_{_{Q}}=13} \\ \\ \\ y_{_{M}}=\dfrac{y_{_{P}}+y_{_{Q}}}{2}\\ \\ 4=\dfrac{2+y_{_{Q}}}{2}\\ \\ 2+y_{_{Q}}=2 \cdot 4\\ \\ 2+y_{_{Q}}=8\\ \\ y_{_{Q}}=8-2\\ \\ \boxed{y_{_{Q}}=6}$


O ponto é $Q\left(13,\,6 \right )$.


b) $P\left(0,\,0 \right )$

$x_{_{M}}=\dfrac{x_{_{P}}+x_{_{Q}}}{2}\\ \\ 5=\dfrac{0+x_{_{Q}}}{2}\\ \\ x_{_{Q}}=2\cdot 5\\ \\ \boxed{x_{_{Q}}=10} \\ \\ \\ y_{_{M}}=\dfrac{y_{_{P}}+y_{_{Q}}}{2}\\ \\ 4=\dfrac{0+y_{_{Q}}}{2}\\ \\ y_{_{Q}}=2 \cdot 4\\ \\ \boxed{y_{_{Q}}=8}$


O ponto é $Q\left(10,\,8 \right )$.