Question

Sejam A=(1,-1) e B=(4,1) vértices do paralelogramo P=(A,B,C,D) . Sabendo que as diagonais de P se cortam no ponto M=(3,2) , encontre os vértices C e D.

Answer 1

Explicação passo-a-passo:

As diagonais de um paralelogramo interceptam-se nos pontos médios

=> Vértice C

As coordenadas do ponto médio são dadas por:

• $\sf x_M=\dfrac{x_A+x_C}{2}$

• $\sf y_M=\dfrac{y_A+y_C}{2}$

Temos:

• $\sf x_M=\dfrac{x_A+x_C}{2}$

$\sf 3=\dfrac{1+x_C}{2}$

$\sf 1+x_C=2\cdot3$

$\sf 1+x_C=6$

$\sf x_C=6-1$

$\sf \red{x_C=5}$

• $\sf y_M=\dfrac{y_A+y_C}{2}$

$\sf 2=\dfrac{-1+y_C}{2}$

$\sf -1+y_C=2\cdot2$

$\sf -1+y_C=4$

$\sf y_C=4+1$

$\sf \red{y_C=5}$

Logo, $\sf \red{C=(5,5)}$

=> Vértice D

As coordenadas do ponto médio são dadas por:

• $\sf x_M=\dfrac{x_B+x_D}{2}$

• $\sf y_M=\dfrac{y_B+y_D}{2}$

Temos:

• $\sf x_M=\dfrac{x_B+x_D}{2}$

$\sf 3=\dfrac{4+x_D}{2}$

$\sf 4+x_D=2\cdot3$

$\sf 4+x_D=6$

$\sf x_D=6-4$

$\sf \red{x_D=2}$

• $\sf y_M=\dfrac{y_B+y_D}{2}$

$\sf 2=\dfrac{1+y_D}{2}$

$\sf 1+y_D=2\cdot2$

$\sf 1+y_D=4$

$\sf y_D=4-1$

$\sf \red{y_D=3}$

Logo, $\sf \red{D=(2,3)}$