Section 2.1 - Rectangular Coordinates and Graphs

Definition 4 (Major Definitions:)

• An ordered pair consists of two components, written inside parentheses.

• Given two real number lines intersecting perpendicularly at \((0,0)\) makes up the rectangular coordinate system.

• The intersection of the two number lines at \((0,0)\) is called the origin.

• The number lines are called axis, the horizontal axis is the \(x\)-axis and the vertical axis is the \(y\)-axis.

• The plane into which the coordinate system is introduced is the coordinate plane, or \(xy\)-plane.

• The \(x\)-axis and \(y\)-axis divide the plane into four regions, or quadrants. Points on the \(x\)-axis or \(y\)-axis belong to no quadrant.

• Let \((x_{1},y_{1})\) and \((x_{2},y_{2})\) be two points on the \(xy\)-plane

  \[\begin{split}\begin{aligned} \Delta x & =x_{2}-x_{1}\\ \Delta y & =y_{2}-y_{1}\\ d & =\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}\\ & =\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\ M & =\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)\end{aligned}\end{split}\]
  where \(\Delta x\) is the change in \(x\), \(\Delta y\) is the change in \(y\), \(d\) is the distance between \((x_{1},y_{1})\) and \((x_{2},y_{2})\), and \(M\) is the midpoint between \((x_{1},y_{1})\) and \((x_{2},y_{2})\).

• The solution to a one variable equation can be represented as \(x=a\) on the \(x\)-axis.

• The solution to a two variable equation can be represented as an ordered on the \(xy\)-plane.

• The graph of an equation is found by plotting ordered pairs that are solutions to the equation, sometimes called constructing a table of values. The intercepts of the graph are good points to plot first. An \(x\)-intercept is a point where the graph intersects the \(x\)-axis. A \(y\)-intercept is a point where the graph intersects the \(y\)-axis.

Example 24

Determine whether the three points are collinear.

• \((0,-7)\), \((-3,5)\), and \((2,-15)\)

Solution:

The distance between \((0,-7)\) and \((-3,5)\) is

\[\begin{align*} d & =\sqrt{(0-(-3))^{2}+(-7-5)^{2}}\\ & =\sqrt{9+144}\\ & =\sqrt{153}\text{ or }3\,\sqrt{17} \end{align*}\]

The distance between \((0,-7)\) and \((2,-15)\) is

\[\begin{align*} d & =\sqrt{(0-2)^{2}+(-7-(-15))^{2}}\\ & =\sqrt{4+64}\\ & =\sqrt{68}\text{ or }2\,\sqrt{17} \end{align*}\]

The distance between \((-3,5)\) and \((2,-15)\) is

\[\begin{align*} d & =\sqrt{(-3-2)^{2}+(5-(-15))^{2}}\\ & =\sqrt{25+400}\\ & =\sqrt{425}\text{ or }5\,\sqrt{17} \end{align*}\]

As we can see \(3\sqrt{17}+2\sqrt{17}=5\sqrt{17}\) which implies the three points are collinear.

• \((-1,-3)\), \((-5,12)\), and \((1,-11)\)

Solution:

The distance from \((-1,-3)\) to \((-5,12)\) is \(\sqrt{241}\approx15.5\).

The distance from \((-1,-3)\) to \((1,-11)\) is \(2\sqrt{17}\approx8.2\).

The distance from \((-5,12)\) to \((1,-11)\) is \(\sqrt{565}\approx23.8\).

Since

\[ \sqrt{241}+2\sqrt{17}\ne\sqrt{565} \]

we say the three points are not collinear.

Example 25

Plot the following from a table of values:

\(x\)	-7	-5	4	6
\(y\)	-5	4	-2	5

Solution:

\(x\)	-3	-2	-1	0
\(y\)	9	4	1	0

Solution:

Given the two variable equation construct a table of values and plot the ordered pairs on the \(xy\)-plane.

1. \(2y-x=-2\)

2. \(y-\sqrt{x}=-3\)

3. \(y=x^{2}\)