What we learn here:

• Cartesian coordinates in a plane
• Ordered pairs
• Formula for the distance between two points in a plane
• Formula of the midpoint of a line segment

Before the 17th century, geometry and algebra were two very distinct branches of mathematics. In 1637 René Descartes, a French mathematician, scientist, and, philosopher, made a huge impact on the development of mathematical knowledge by unifying these two branches of mathematics. His approach is now called analytic geometry, the key feature of which is the use of a coordinate system. By means of coordinate systems, we can employ algebraic methods to study geometry and can geometrically represent algebraic equations. Analytic geometry is now widely used in mathematics, science, and engineering.

## Points

In Section 1.3 “Sets of Numbers“, we showed how the real number system can be used to assign coordinates to points on a line.

#### Summary of assigning coordinates to points on a line

On a straight line, we choose a definite length as a unit distance and take a fixed point $O$ from which to measure distances. Any point on this line represents a real number. Points on the right of $O$ represent positive numbers and points on the left of $O$ represent negative ones. The point $O$ is called the origin and represents the number zero. To determine the position of a point on such a line, only one number is sufficient, namely the distance of the point, right or left, from the origin.

In the plane, to determine the position of a point $P$, we need two real numbers, namely a pair of numbers $(a,b)$. We draw two perpendicular straight lines, label them as the $x$ and $y$ axes, designate their intersection as the origin, choose the same unit distance on both axes, and lay off distances from the origin along the axes. The distances to the right along the $x$-axis are taken as positive and those to the left as negative. In the same way, distances upward along the $y$-axis are taken as positive and those downward as negative.

In this case, the position of a point $P$ in the plane can be clearly indicated by drawing two straight lines perpendicular to the axes, intersecting the $x$ and $y$ axes at $M$ and $N$, respectively. Two signed distances $a=\overline{OM}$ and $b=\overline{ON}$ are called respectively the abscissa (or the $\boldsymbol{x}$ component) of $P$ and the ordinate (or the $\boldsymbol{y}$ component) of $P$ and together known as the rectangular coordinates or the Cartesian coordinates (in honor of René Descartes) of the point. In this case, we write $P(a,b)$ (see the following figure).

• The $x$ and $y$ axes divide all of the plane into four regions called quadrants. The quadrants are numbered I, II, III, and IV (called first, second, third, and fourth quadrants) as in the above figure.

Because in general $(a,b)\neq (b,a)$, the pair $(a,b)$ is often referred to as an ordered pair.

#### more explanation about ordered pairs

For example, to locate the point with coordinates $(4,-3)$, we go out 4 units on the $x$-axis and $-3$ units on the $y$-axis. But to locate the point with coordinates $(-3,4)$, we go out $-3$ units on the $x$-axis and 4 units on the $y$-axis. Obviously, a point with coordinates $(4,-3)$ is different from a point with coordinates $(-3,4)$ (see the following figure). • Instead of “the point with coordinates $(a,b)$,”  we often simply say “the point $(a,b)$.”

Don't get confused about the symbol (a, b)

The symbol $(a,b)$ was previously used to denote an open interval, but in this section, such a symbol represents a point. Whether the symbol $(a,b)$ represents an interval, a point, or some other mathematical object is often clear and can be understood from the context.

• The set of all ordered pairs is denoted by $\mathbb{R\times R}$ or $\mathbb{R}^{2}$

$\mathbb{R}^{2}=\{(a,b)|\ a,b\in\mathbb{R}\}.$

Example

Describe and sketch the regions given by the following sets

(a) $\left\{(x,y)|\ y\le0\right\}$

(b) $\{(x,y)|\ x=2\}$

(c) $\{(x,y)|\ y=3\}$

(d) $\{(x,y)|\ x\geq2,y<3\}$

(e) $\{(x,y)|\ |y|<3\}$

Solution

(a) The set of all points whose $y$-coordinates are negative or zero consists of the $x$-axis and all points below the $x$-axis as indicated by the shaded region in the following figure. (b) The sets of all points with $x$-coordinate 2 is a vertical line 2 units to the right of the $y$-axis (see the following figure). (c) The sets of all points with $y$-coordinate 3 is a horizontal line 3 units above the $x$-axis (see the following figure) (d) The sets of all points with $x$-coordinate greater or equal to 2 and $y$-coordinate less than 3 consists of points that lie on the vertical line $x=2$ or to the right of it and below the line $y=3$, as indicated by the shaded region in the following figure. The dashed line indicates the points on that do not lie in the set. (e) Because $|y|<3$ is equivalent to $-3<y<3$, this set consists of all points between the vertical lines $y=3$ (horizontal line 3 units above the $x$-axis) and $y=-3$ (horizontal line 3 units below the $x$-axis). ### Distance Between Points

Because we used the same unit of length on both axes, we can apply geometry to express the distance between two points in terms of their coordinates. Consider two points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ as in the following figure. If we apply the Pythagorean theorem to the right triangle $PRQ$, we realize the distance between $P$ and $Q$, denoted by $d(P,Q)$ or $|PQ|$ is given by:

$d(P,Q)=|PQ|={\sqrt{|x_{2}-x_{1}|^{2}+|y_{2}-y_{1}|^{2}}}={\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}\tag{i}$ Example

Find the distance between the points $P(1,-5)$ and $Q(-1,3)$.

Solution

Using the distance formula

\begin{align*}
d(P,Q)=|PQ| & =\sqrt{(-1-1)^{2}+(3-(-5))^{2}}\\
& =\sqrt{(-2)^{2}+8^{2}}\\
& =\sqrt{68}
\end{align*}

### The Midpoint Formula

The coordinates of the midpoint $M(x_{m},y_{m})$ of the line segment joining $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$ is

$(x_{m},y_{m})=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$

• Remark that $|PM|=|MQ|$ (or $d(P,M)=d(M,Q)$) Example

Find the midpoint of the line segment with endpoints $(-3,1.5)$ and $(7,-4.5)$.

Solution

Using the formula, the coordinates of the midpoint are

\begin{align*}
\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right) & =\left(\frac{-3+7}{2},\frac{1.5-4.5}{2}\right)\\
& =\left(2,-1.5\right)
\end{align*}