Chapter 8.6, Problem 48E

Solution

To find: The midpoint of $AM$ where $M$ is midpoint of $BC$ .

The midpoint of $AM$ is $\left(0,\frac{3}{2},\frac{3}{2}\right)$ where $M\left(1,1,-1\right)$ is midpoint of $BC$ .

Given information:

The given points are $A\left(-1,2,4\right)$ , $B\left(0,6,-3\right)$ and $C\left(2,-4,1\right)$ .

Formula used:

Midpoint formula: The midpoint $M$ of the line segment $PQ$ with endpoints $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ is $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$ .

Calculation:

Consider the points $A\left(-1,2,4\right)$ , $B\left(0,6,-3\right)$ and $C\left(2,-4,1\right)$ .

  $M$ is the midpoint of $BC$ , then

Using the formula $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substitute $0$ for $x_1$ , $6$ for $y_1$ , $-3$ for $z_1$ , $2$ for $x_2$ , $-4$ for $y_2$ and $1$ for $z_2$ .

  $\begin{array}{l}M=\left(\frac{0+2}{2},\frac{6+\left(-4\right)}{2},\frac{-3+1}{2}\right)\\ M=\left(\frac{2}{2},\frac{2}{2},\frac{-2}{2}\right)\\ M=\left(1,1,-1\right)\end{array}$

Thus the midpoint of $BC$ is $M=\left(1,1,-1\right)$

Let $N$ be the midpoint of $AM$ .

Again using the formula $N=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

Substitute $-1$ for $x_1$ , $2$ for $y_1$ , $4$ for $z_1$ , $1$ for $x_2$ , $1$ for $y_2$ and $-1$ for $z_2$ .

  $\begin{array}{l}N=\left(\frac{-1+1}{2},\frac{2+1}{2},\frac{4-1}{2}\right)\\ N=\left(\frac{0}{2},\frac{3}{2},\frac{3}{2}\right)\\ N=\left(0,\frac{3}{2},\frac{3}{2}\right)\end{array}$

Thus, the midpoint of $AM$ is $\left(0,\frac{3}{2},\frac{3}{2}\right)$ where $M\left(1,1,-1\right)$ is midpoint of $BC$ .