Chapter 10.5, Problem 57AYU

To determine

To find:

The formula of the half midpoint

Answer to Problem 57AYU

  $M_1=x_1+\frac{1}{2}\left(x_2-x_1\right)$

Explanation of Solution

Given:

The point is $\frac{1}{2}$ of the distance from the point $P\left(x_1,y_1\right)$ to the point $Q\left(x_2,y_2\right)$

Concept used:

The midpoint $M$ of the line segment from $P\left(x_1,y_1\right),Q\left(x_2,y_2\right)$ is

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

Calculation:

The line through $P$ and $Q$ parallel to the y-axis at $A_1\left(x_1,0\right)$ and $A_1\left(x_2,0\right)$ .

From plane geometry the line through the midpoint $M$ parallel to the y-axis bisects the segment $A_1{A}_2$ at point $M_1$

If $x_1<x_2$ then $x_2-x_1>0$ and hence $d\left(A_1,A_2\right)=x_2-x_1$

Since $M_1$ is halfway from $A_1$ to $A_2$

The x-coordinate of $M_1$ is equal to the x-coordinate of $A_1$ plus one half the distance from $A_1$ to $A_2$

That is

  $x-$ Coordinate of $M_1=x_1+\frac{1}{2}\left(x_2-x_1\right)$