Question

O ponto M é o ponto médio do segmento AB. Determine o ponto B em cada caso.


A) A (4,-8) e M (0,-1)

B) A ( -2, 5) e M ( -2, 6)

C) A ( -1, 5) e M (3,-8)​

Answer 1

Explicação passo-a-passo:

Temos que:

$x_M=\dfrac{x_A+x_B}{2}$

$y_M=\dfrac{y_A+y_B}{2}$

a) A(4, -8) e M(0, -1)

$\bullet~x_M=\dfrac{x_A+x_B}{2}$

$0=\dfrac{4+x_B}{2}$

$4+x_B=2\cdot0$

$4+x_B=0$

$x_B=-4$

$\bullet~y_M=\dfrac{y_A+y_B}{2}$

$-1=\dfrac{-8+y_B}{2}$

$-8+y_B=2\cdot(-1)$

$-8+y_B=-2$

$y_B=-2+8$

$y_B=6$

Logo, $B(-4,6)$

b) A(-2, 5) e M(-2, 6)

$\bullet~x_M=\dfrac{x_A+x_B}{2}$

$-2=\dfrac{-2+x_B}{2}$

$-2+x_B=2\cdot(-2)$

$-2+x_B=-4$

$x_B=-4+2$

$x_B=-2$

$\bullet~y_M=\dfrac{y_A+y_B}{2}$

$6=\dfrac{5+y_B}{2}$

$5+y_B=2\cdot6$

$5+y_B=12$

$y_B=12-5$

$y_B=7$

Logo, $B(-2,6)$

c) A(-1, 5) e M(3, -8)

$\bullet~x_M=\dfrac{x_A+x_B}{2}$

$3=\dfrac{-1+x_B}{2}$

$-1+x_B=2\cdot3$

$-1+x_B=6$

$x_B=6+1$

$x_B=7$

$\bullet~y_M=\dfrac{y_A+y_B}{2}$

$-8=\dfrac{5+y_B}{2}$

$5+y_B=2\cdot(-8)$

$5+y_B=-16$

$y_B=-16-5$

$y_B=-21$

Logo, $B(7,-21)$