# 2.1 The rectangular coordinate systems and graphs  (Page 5/21)

 Page 5 / 21

## Using the midpoint formula

When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula    . Given the endpoints of a line segment, $\text{\hspace{0.17em}}\left({x}_{1},{y}_{1}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left({x}_{2},{y}_{2}\right),$ the midpoint formula states how to find the coordinates of the midpoint $\text{\hspace{0.17em}}M.$

$M=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$

A graphical view of a midpoint is shown in [link] . Notice that the line segments on either side of the midpoint are congruent.

## Finding the midpoint of the line segment

Find the midpoint of the line segment with the endpoints $\text{\hspace{0.17em}}\left(7,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(9,5\right).$

Use the formula to find the midpoint of the line segment.

$\begin{array}{l}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)=\left(\frac{7+9}{2},\frac{-2+5}{2}\right)\hfill \\ \phantom{\rule{6.5em}{0ex}}=\left(8,\frac{3}{2}\right)\hfill \end{array}$

Find the midpoint of the line segment with endpoints $\text{\hspace{0.17em}}\left(-2,-1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-8,6\right).$

$\left(-5,\frac{5}{2}\right)$

## Finding the center of a circle

The diameter of a circle has endpoints $\text{\hspace{0.17em}}\left(-1,-4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,-4\right).\text{\hspace{0.17em}}$ Find the center of the circle.

The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point.

$\begin{array}{c}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\\ \left(\frac{-1+5}{2},\frac{-4-4}{2}\right)=\left(\frac{4}{2},-\frac{8}{2}\right)=\left(2,-4\right)\end{array}$

Access these online resources for additional instruction and practice with the Cartesian coordinate system.

## Key concepts

• We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x- axis and displacement from the y- axis. See [link] .
• An equation can be graphed in the plane by creating a table of values and plotting points. See [link] .
• Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y= _____. See [link] .
• Finding the x- and y- intercepts can define the graph of a line. These are the points where the graph crosses the axes. See [link] .
• The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See [link] and [link] .
• The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x -coordinates and the sum of the y -coordinates of the endpoints by 2. See [link] and [link] .

## Verbal

Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain.

Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.

Describe the process for finding the x- intercept and the y -intercept of a graph algebraically.

Describe in your own words what the y -intercept of a graph is.

The y -intercept is the point where the graph crosses the y -axis.

When using the distance formula $\text{\hspace{0.17em}}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}},$ explain the correct order of operations that are to be performed to obtain the correct answer.

## Algebraic

For each of the following exercises, find the x -intercept and the y -intercept without graphing. Write the coordinates of each intercept.

$y=-3x+6$

The x- intercept is $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and the y -intercept is $\text{\hspace{0.17em}}\left(0,6\right).$

$4y=2x-1$

$3x-2y=6$

The x- intercept is $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and the y -intercept is $\text{\hspace{0.17em}}\left(0,-3\right).$

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