Question

Identify the distance between points (3,8, 0) and (-2,9,-4), and identify the midpoint of the

Answer 1

Answer: C) d-6.5 units: M(0.5, 8.5.-2)

Step-by-step explanation:

Distance between points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$  :

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The midpoint of the line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by :-

$(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2})$

The given points : (3,8, 0) and (-2,9,-4)

Distance between points (3,8, 0) and (-2,9,-4):

$D=\sqrt{(-2-3)^2+(9-8)^2+(-4-0)^2}\\\\=\sqrt{25+1+16}\\\\=\sqrt{42}=6.48074069841\approx6.5\text{units}$

The midpoint of the line joining (3,8, 0) and (-2,9,-4) :

$(\dfrac{3+(-2)}{2},\dfrac{8+9}{2},\dfrac{0+(-4)}{2})=(\dfrac{1}{2},\dfrac{17}{2},-2)\\\\=(0.5, 8.5.-2)$

Hence, the correct answer is C) d-6.5 units: M(0.5, 8.5.-2).