Chapter 8.3, Problem 41WE

To determine

To show: That the point $M\left(6,0\right)$ lie on the diagonal joined by the coordinates $A\left(3,5\right)$ and $C\left(9,-5\right)$ , and on the diagonal joined by $B\left(11,3\right)$ and $D\left(1,-3\right)$ .

Explanation of Solution

Given information:

  $ABCD$ is a square where the points are:

  $A\left(3,5\right),B\left(11,3\right),C\left(9,-5\right),D\left(1,-3\right)$

And the sides of the square are:

  $AB,BC,CD,DA$

Formula used:

The mid-point formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Consider the vertices of a square:

  $A\left(3,5\right),B\left(11,3\right),C\left(9,-5\right),D\left(1,-3\right)$

Now first find the mid-point between the points $A$ and $M$ .

Recall that, the mid-point formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

The points of $A$ and $M$ are:

  $A\left(3,5\right),M\left(6,0\right)$

Here,

  $\begin{array}{l}{x}_1=3,x_2=6\\ y_1=5,y_2=0\end{array}$

Apply these values in the formula of mid-point.

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{0-5}{6-3}\\ m=\frac{-5}{3}\end{array}$

Now find the mid-point between the points $C$ and $M$ .

Recall that, the mid-point formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

The points of $C$ and $M$ are:

  $C\left(9,-5\right),M\left(6,0\right)$

Here,

  $\begin{array}{l}{x}_1=9,x_2=6\\ y_1=-5,y_2=0\end{array}$

Apply these values in the formula of mid-point.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{0-\left(-5\right)}{6-9}\\ m=\frac{5}{-3}\\ m=-\frac{5}{3}\end{array}$

Here the mid-point of $C$ and $M$ is same as of the $A$ and $M$ .

So the point $M$ lies on the diagonal lies from the points $A$ to point $C$ .

Now find the mid-point between the points $B$ and $M$ .

Recall that, the mid-point formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

The points of $C$ and $M$ are:

  $B\left(11,3\right),M\left(6,0\right)$

Here,

  $\begin{array}{l}{x}_1=11,x_2=6\\ y_1=3,y_2=0\end{array}$

Apply these values in the formula of mid-point.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{0-3}{6-11}\\ m=\frac{-3}{-5}\\ m=\frac{3}{5}\end{array}$

Now find the mid-point between the points $D$ and $M$ .

Recall that, the mid-point formula:

  $m=\frac{y_2-y_1}{x_2-x_1}$

The points of $C$ and $M$ are:

  $D\left(1,-3\right),M\left(6,0\right)$

Here,

  $\begin{array}{l}{x}_1=1,x_2=6\\ y_1=-3,y_2=0\end{array}$

Apply these values in the formula of mid-point.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{0-\left(-3\right)}{6-1}\\ m=\frac{3}{5}\end{array}$

Here the mid-point of $C$ and $M$ is same as of the $A$ and $M$ .

So the point $M$ lies on the diagonal lies from the points $A$ to point $C$ .

Thus, the point $M\left(6,0\right)$ lies on the diagonal joining $A$ and $C$ , and $B$ and $D$ of a square.