Chapter 8.1, Problem 37E

Solution

To find: two points not on the lines $x=4$ or $y=2$ such that the midpoint of the line segment joining the points is $\left(4,2\right)$ .

  $\left(6,4\right)$ and $\left(2,0\right)$

Given information:

The lines $x=4$ or $y=2$ are given.

Formula Used:

The Midpoint Formula

The midpoint formula shown below, gives the midpoint of the line segment joining the points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ .

  $M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

  

Explanation:

Let $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ be the two required points, for which the midpoint of the line segment joining the points is $\left(4,2\right)$ .

Then, by the midpoint formula,

  $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(4,2\right)$

On equating the coordinates two equations are obtained.

  $\begin{array}{l}\frac{x_1+x_2}{2}=4\\ \frac{y_1+y_2}{2}=2\end{array}$

This gives that,

  $\begin{array}{l}{x}_1+x_2=8\\ y_1+y_2=4\end{array}$

Now, choose any numbers that are not $x=4$ and $y=2$ . For example, let $x_1=6$ and $y_1=4$

  $\begin{array}{c}6+x_2=8\\ x_2=8-6\\ x_2=2\end{array}$

Similarly, the value of other variable is:

  $\begin{array}{c}4+y_2=4\\ y_2=4-4\\ y_2=0\end{array}$

So, one possible solution is $\left(6,4\right)$ and $\left(2,0\right)$ .