Q41. - Algebra 2

Question

CRITICAL THNKING 

Verify the Midpoint Formula. (Hint: You must show that

the formula gives the coordinates of a point on the line through the given

endpoints and that the point is equidistant from the endpoints.)

Step-by-Step Solution

Answer

We have proved the midpoint formula in calculation section.

Step 1. Given information

We have to verify the midpoint theorem

Step 2. Concept used

The midpoint theorem is that the formula gives the coordinates of a point on the line through the given endpoints and that the point is equidistant from the endpoints.

Step 3. Calculation

We assumed a line such as-

The slope of the line through $(x_1,y_1)$and$\left(x_2,y_2\right)$  is $\frac{y_2-y_1}{x_2-x_1}$ and the point-slope form of the equation of the line is $$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$.  Substitute $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ into this equation. 

The left side is$\frac{y_1+y_2}{2}-y_1\&\frac{y_2-y_1}{2}$ . The right side is $\frac{y_2-y_1}{x_2-x_1}\left(\frac{x_1+x_2}{2}-x_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(\frac{x_2-x_1}{2}\right)$  or $\frac{y_2-y_1}{2}$. Therefore, the point with coordinates $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$  lies on the line through $\left(x_1,y_1\right)$ and$\left(x_2,y_2\right)$.

The distance from $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$to $\left(x_1,y_1\right)$is$\sqrt{{\left(x_1-\frac{x_1+x_2}{2}\right)}^2+{\left(y_1-\frac{y_1+y_2}{2}\right)}^2}$ or

$\sqrt{{\left(\frac{x_1-x_2}{2}\right)}^2+{\left(\frac{y_1-y_2}{2}\right)}^2}$.

 

The distance from $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$  to $(x_2,y_2)$ is

$\begin{array}{l}\sqrt{{\left(x_2-\frac{x_1+x_2}{2}\right)}^2+{\left(y_2-\frac{y_1+y_2}{2}\right)}^2}\\ =\sqrt{{\left(\frac{x_2-x_1}{2}\right)}^2+{\left(\frac{y_2-y_1}{2}\right)}^2}\end{array}$

 or $\sqrt{{\left(\frac{x_1-x_2}{2}\right)}^2+{\left(\frac{y_1-y_2}{2}\right)}^2}$

the distance from$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ to $\left(x_2,y_2\right)$ is

$\begin{array}{l}\sqrt{{\left(x_2-\frac{x_1+x_2}{2}\right)}^2+{\left(y_2-\frac{y_1+y_2}{2}\right)}^2}\\ =\sqrt{{\left(\frac{x_2-x_1}{2}\right)}^2+{\left(\frac{y_2-y_1}{2}\right)}^2}\end{array}$

or $\sqrt{{\left(\frac{x_1-x_2}{2}\right)}^2+{\left(\frac{y_1-y_2}{2}\right)}^2}$ 

therefore the points with coordinates $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ is equidistant from $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$.

Hence proved.