Polygons in a coordinate plane will make it easy to find the distance between the points of the polygon. It also helps in finding the perimeter and area of the polygon. 

We need to know that a horizontal line goes straight from left to right, a vertical line goes from up to down and the distance is how far apart the two points are. 

If we have a straight horizontal line then the $y$ coordinates will remain the same. Similarly, for a straight vertical line the $x$ coordinates will remain the same. 

If the vertices of a quadrilateral are $A (2, 4)$, $B (3, 9)$, $C (7, 8)$ and $D (8, 1)$; draw the quadrilateral in a coordinate plane.

Here, we see that all the coordinates are positive. So, we just need the first quadrant. We also should know that for a given coordinate the x-axis comes first and the y-axis comes second. So, for the vertex $A (2, 4), 2$ will be the x-coordinate and $4$ the y-coordinate. We will go over $2$ and up $4$ for the point $A (2, 4)$.
 
Similarly we can plot also the coordinates $B (3, 9)$, $C (7, 8)$ and $D (8, 1)$.

Once we are done with plotting the coordinates we have to join the dots with a straight line. We need to go sequentially from $A$ to $B$, $B$ to $C$, $C$ to $D$ and $D$ to $A$ as shown in the following figure:

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We can classify a polygon in the coordinate plane by determining the length of the side and checking if the lengths are parallel (i.e. by determining their slopes).

To classify a triangle we should first know the different types of triangles. 

We can classify triangles by angles: acute, obtuse or right

We can classify triangles by sides: scalene (all sides have different lengths), isosceles (two sides are equal) or equilateral (all sides are equal) 

Let us take an example:
 
Is $\Delta$ $ABC$ scalene, isosceles or equilateral?

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To determine if the given triangle is scalene, isosceles or equilateral we need to find the length of the sides. The length of a side will be the distance between the two points. 

We know the coordinates of the given points are: 

$A$ =$(0,1)$, $B$ =$(4,4)$  and $C$ = $(7,0)$

Now, we can find distance between the points by using distance formula.

$d_{AB}$ = $\sqrt{((4-0)^2+(4-1)^2)}$ = $\sqrt{25}$ = 5

$d_{BC}$ = $\sqrt{((7-4)^2+(0-4)^2)}$ = $\sqrt{25}$ = 5

$d_{AC}$ = $\sqrt{((7-0)^2+(0-1)^2)}$ = $\sqrt{50}$ = 5$\sqrt{2}$

So, we get two sides of equal length and hence this will be an isosceles triangle. 

Let us look at another example

Is parallelogram $ABCD$ a rhombus? Explain. 

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We should know that a rhombus has all sides congruent and diagonals perpendicular. 

To check if all the sides are congruent we need to find the distance of the sides. 

The coordinates are: 

$A$ = $(-2,0)$, $B$ = $(0,4)$,$C$ =$(4,5)$ and $D$ = $(2,1)$

$d_{AB}$ = $\sqrt{((0+2)^2+(4-0)^2)}$ = $\sqrt{20}$ = 2$\sqrt{5}$

$d_{AD}$ = $\sqrt{((2+2)^2+(1-0)^2)}$ = $\sqrt{17}$  

Since, the sides $AB$ and $AD$ are not equal so we can say that this is not a rhombus. No further calculations are needed.
 
Similarly, if we had to check if a parallelogram is a rectangle we need to check weather the diagonals are congruent.

To check weather a parallelogram is a square we need to check if all properties of rhombuses and rectangles hold true.
Question 1:

Draw the polygon with the given vertices in a coordinate plane. 

a) $A (0, 0),\ B (5, 7),\ C (7, 4)$

Solution:
 
To draw the given polygon we can mark the given points and join them.

image

b) $F(1, 3),\ G(3, 6),\ H(5, 6),\ I(3, 3)$

Solution:

To draw the polygon with given vertices we need to plot the points in coordinate axis and join them.

image

c) $P (1,4),\ Q(3,5),\ R (7,3 ),\ S(6,$ $\frac{1}{2})$, $T(2,$ $\frac{1}{2})$
 
Solution: 

We can draw the given polygon by drawing the coordinate points and joining them. 

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Question 2:

Is the given parallelogram $ABCD$ a rectangle? Explain.

image

Solution:

To check if the given parallelogram is a rectangle we need to check if their diagonals are equal. 

First, we need check the coordinates of the points.

$A$ =$(-5,1)$, $B$ = $(5,6)$, $C$ =$(8,1)$  and $D$ = $(-2,-4)$

Length of the diagonal AC will be:

$d_{AC}$ = $\sqrt{((-5 -8)^2+(1-1)^2)}$ = $\sqrt{((-13)^2)}$ = $13$

Length of the diagonal BD will be:

$d_{BD}$ = $\sqrt{((5+2)^2+(6+4)^2)}$ = $\sqrt{(7^2 + 10^2)}$ = $\sqrt{149}$ $\approx$ 12.21

Since, the lengths of the diagonals are not equal so the given parallelogram will not be a rectangle.