Graphing Polygons on the Coordinate Plane | Grade 8 Math | LevelKite

You already know how to plot one ordered pair $(x, y)$ : the $x$ -coordinate gives the position left or right of the origin $(0, 0)$ , and the $y$ -coordinate gives the position up or down. Now you will use several ordered pairs together to draw a whole figure.

A polygon is a closed figure made of three or more straight sides that meet only at their endpoints. Each point where two sides meet is a vertex (plural: vertices).

On the coordinate plane, you describe a polygon by listing its vertices, such as triangle $PQR$ with $P(1, 1)$ , $Q(5, 1)$ and $R(3, 4)$ . Every list in this lesson names the vertices in order around the figure, so the list tells you two things:

Where each vertex is: its exact ordered pair.
Which vertices are joined: each vertex to the next one in the list, and the last one back to the first.

For triangle $PQR$ , the sides are $\overline{PQ}$ , $\overline{QR}$ and $\overline{RP}$ , where $\overline{PQ}$ means the straight side from $P$ to $Q$ . To graph a figure means to plot every vertex and draw exactly those sides.

Key concept

Coordinates and order define the figure

When a list gives a polygon's vertices as exact ordered pairs, in order around the figure, you graph the polygon in three steps:

Plot every vertex at its exact coordinates.
Join each vertex to the next one in the list with a straight side.
Join the last vertex back to the first to close the figure.

The coordinates and the order of the list decide the figure, never the way a drawing looks. Any fact you state about the figure must come from its coordinates or from a stated definition. Every grid in this lesson uses 1 unit per grid step on both axes.

Worked example

Graph a four-sided figure

Graph quadrilateral $ABCD$ (a polygon with four sides) with vertices $A(-3, 2)$ , $B(4, 2)$ , $C(4, -2)$ and $D(-3, -2)$ , listed in order around the figure. Each grid step is 1 unit on both axes.

Plot vertex A at its exact ordered pair. The x-coordinate is -3, a negative number, so go 3 units left of the origin. The y-coordinate is 2, a positive number, so go 2 units up. The sign tells you the direction; the distance you move is a count of units, so it is never negative.
$A(-3, 2)$
Plot the other three vertices the same way, because every vertex must sit at its exact coordinates. B is 4 units right of the origin and 2 up, C is 4 units right and 2 down, and D is 3 units left and 2 down.
$B(4, 2), \quad C(4, -2), \quad D(-3, -2)$
Join each vertex to the next one in the list with a straight side: A to B, B to C, then C to D. Vertices that are next to each other in the list are the endpoints of a side, so the list gives three sides so far.
$\overline{AB}, \quad \overline{BC}, \quad \overline{CD}$
Join the last vertex, D, back to the first vertex, A. A polygon is a closed figure, so it needs this fourth side.
$\overline{DA}$

Answer

The graph of quadrilateral $ABCD$ has its vertices at $A(-3, 2)$ , $B(4, 2)$ , $C(4, -2)$ and $D(-3, -2)$ and has the four sides $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ and $\overline{DA}$ , as the figure below shows.

Point A is at

(-3, 2)

, Point B at

(4, 2)

, Point C at

(4, -2)

and Point D at

(-3, -2)

; the sides join them in that order and back to Point A.

The figure above matches the list exactly: Point A is at $(-3, 2)$ , Point B at $(4, 2)$ , Point C at $(4, -2)$ and Point D at $(-3, -2)$ .

Now check the sides with the coordinates instead of trusting your eyes:

$A$ and $B$ both have $y$ -coordinate $2$ , so both lie on the grid line 2 units above the $x$ -axis, and the straight side $\overline{AB}$ runs along that line. The side is horizontal, parallel to the $x$ -axis. $\overline{CD}$ is horizontal too, because $C$ and $D$ both have $y$ -coordinate $-2$ .
$B$ and $C$ both have $x$ -coordinate $4$ , so both lie on the grid line 4 units to the right of the $y$ -axis, and $\overline{BC}$ runs along that line. The side is vertical, parallel to the $y$ -axis. $\overline{DA}$ is vertical too, because $D$ and $A$ both have $x$ -coordinate $-3$ .

The $x$ -axis and $y$ -axis meet at a right angle, so a horizontal side and a vertical side do too. At every vertex of $ABCD$ a horizontal side meets a vertical side, so all four angles are right angles, and a quadrilateral with four right angles is a rectangle. The coordinates prove this; the drawing only agrees.

Worked example

Let the list choose the sides

Graph quadrilateral $EFGH$ with vertices $E(-4, -3)$ , $F(0, -1)$ , $G(4, -3)$ and $H(0, 5)$ , listed in order around the figure. Each grid step is 1 unit on both axes.

Plot all four vertices at their exact ordered pairs. E is 4 units left of the origin and 3 down; G is 4 units right and 3 down. F and H have x-coordinate 0, so they are neither left nor right of the origin and lie on the y-axis: F is 1 unit below the origin and H is 5 units above it.
$E(-4, -3), \quad F(0, -1), \quad G(4, -3), \quad H(0, 5)$
Join each vertex to the next one in the list with a straight side: E to F, F to G, then G to H. Do not jump to whichever point looks closest, because the list decides which vertices are joined.
$\overline{EF}, \quad \overline{FG}, \quad \overline{GH}$
Close the figure by joining the last vertex, H, back to the first vertex, E, because a polygon is a closed figure.
$\overline{HE}$
Check the sides against the list. F and H share the x-coordinate 0, and E and G share the y-coordinate -3, but neither pair is next to each other in the list. So no side joins F to H, and no side joins E to G.

Answer

The graph of quadrilateral $EFGH$ has its vertices at $E(-4, -3)$ , $F(0, -1)$ , $G(4, -3)$ and $H(0, 5)$ and has the four sides $\overline{EF}$ , $\overline{FG}$ , $\overline{GH}$ and $\overline{HE}$ , as the figure below shows.

Point E is at

(-4, -3)

, Point F at

(0, -1)

, Point G at

(4, -3)

and Point H at

(0, 5)

; the sides join them in the listed order and back to Point E.

The mistaken order

E

,

F

,

H

,

G

: one side now runs from Point F at

(0, -1)

to Point H at

(0, 5)

, and another from Point G at

(4, -3)

to Point E at

(-4, -3)

.

The order you join decides the sides

Order joined	Sides drawn	Same figure as EFGH?
$E$ , $F$ , $G$ , $H$ (as listed)	$\overline{EF}$ , $\overline{FG}$ , $\overline{GH}$ , $\overline{HE}$	Yes
$E$ , $H$ , $G$ , $F$ (the other way around)	$\overline{EH}$ , $\overline{HG}$ , $\overline{GF}$ , $\overline{FE}$	Yes, the same four sides
$E$ , $F$ , $H$ , $G$	$\overline{EF}$ , $\overline{FH}$ , $\overline{HG}$ , $\overline{GE}$	No, two sides are different

The table shows when the order matters. Going around the other way, $E$ , $H$ , $G$ , $F$ , keeps the same pairs of vertices next to each other, so it draws the same four sides ( $\overline{EH}$ and $\overline{HE}$ name the same side). The order $E$ , $F$ , $H$ , $G$ changes which vertices are next to each other, so it draws a different figure.

Before you call a graph finished, check it against the list, not against how it looks:

Read each plotted vertex back from the grid. Its ordered pair must match the list exactly. When every coordinate is an integer and each grid step is 1 unit, as in this lesson, every vertex lands where two grid lines cross.
Count the sides. A polygon has as many sides as vertices.
Confirm that each side joins two vertices that are next to each other in the list, counting the last vertex and the first as a pair.
Where two joined vertices share a $y$ -coordinate, the side must be horizontal; where they share an $x$ -coordinate, it must be vertical.

Takeaway: A figure on the coordinate plane is fixed by the list of its vertices, taken in order. Plot each vertex at its exact ordered pair, join the vertices in the listed order, and join the last vertex back to the first. Whatever you say about the figure should come from its coordinates and stated definitions, not from how the drawing looks.

Next, you will find the distance between two points on the same horizontal or vertical line and solve word problems on the coordinate plane.