# Geometry for Elementary School/Rectangular coordinate system

 Geometry for Elementary School Coordinates Rectangular coordinate system Polar coordinate system

The rectangular coordinate system, or Cartesian coordinate system, is a systematic way of locating objects. It was invented by a mathematician called Descartes. It is similar, though not exactly the same as, the coordinate systems in geography or computer literacy.

## Basic concept

A summary of basic Cartesian plane-related terms

A rectangular coordinate plane, or Cartesian plane, is made of two axes, the x-axis and the y-axis. The x-axis is the horizontal axis, while the y-axis is the vertical one. The point of intersection of these two axes is called the origin, and is always named O. On the x-axis and y-axis is the scale, which is very similar to those in statistical diagrams, only there are negative numbers. In fact, they look like number lines, one horizontal and one vertical, with the '0' points overlapping.

• The x-axis: (x or -x, 0)
• The y-axis: (0, y or -y)
• The origin: (1, 1)

When determining the ordered pair of a point, remember that the scale may not necessarily be one-to-one. That means the numbers don't have to go 0, 1, 2, 3, 4. They can also go 0, 2, 4, 6, 8 or even 5, 10, 15, 20.

If you are absolutely sure that two points form a horizontal of vertical line, the definition of horizontal being parallel to the x-axis and the definition of vertical being parallel to the y-axis, you can find the one or both of the coordinates of one point with reference to another, without the help of the scale. For example, if P (3, 4) is on the same vertical line as Q (3, y), then y = 4.

## Determining length and area

We will introduce the next part with a complex problem (see the figure on the right).
Given:

• XD//YE

Find:

• The length of QP
• The sum of all shaded areas
• The coordinates of I if F: (-2, -3)

It looks a bit complicated at first, so let us start from the line on the first quadrant, QP, which appears to be rather easy. We will work out all the unknowns if we can as well as solve the first bulleted goal. First, let's take a look at the coordinates of P. They are (1, 2). As P and Q lie on the same horizontal line, their y-coordinate must be the same. Therefore, the coordinates of Q must be (3, 2). In other words, z = 2. To find the length of PQ, all we need to do is to find the difference between their x-coordinates; in this case, 3 - 1 = 2 units. Therefore, QP = 2 units. In mathematical representation, that would be:

{\displaystyle {\begin{aligned}\because {\text{The }}y{\text{-coordinate of }}P=2\\\therefore z=2\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The length of }}\ PQ&=3-1\\&=2{\text{ units}}\end{aligned}}}

Good! Now that we have solved the first goal, let's move on to the little triangle on the second quadrant. Again, we'll work out all the unknowns. First, let's look at the x-coordinate of C. we already know that z = 2, so -z must be -2, and so is the x-coordinate of C. What about the y-coordinate? we know that B and C lie on the same horizontal line, so their y-coordinates must be the same. Therefore, the y-coordinate of C must be 1 also, and y = 1. The x-coordinate of B, called x, is also missing. As A and B are on the same vertical line, their x-coordinate must be the same. So, we can deduce that x = -4. In mathematical expressions:

{\displaystyle {\begin{aligned}{\text{The }}\ x{\text{-coordinate of }}\ C&=-z\\&=1\end{aligned}}}

{\displaystyle {\begin{aligned}\because {\text{The }}y{\text{-coordinate of }}B=1\\\therefore y=1\end{aligned}}}

{\displaystyle {\begin{aligned}\because {\text{The }}x{\text{-coordinate of }}A=-4\\\therefore x=-4\end{aligned}}}

Next, we will find out the area of the triangle, ABC. This is going to prove to be very helpful when finding the total area of the shaded regions. Before we do that, we must find out the height and base of the triangle. Do you still remember the formula for the area of triangles? It is height times base over two. Let's choose BC as the base of the triangle. As AB and BC are perpendicular to each other, AB can be the height. As BC is horizontal, the length of BC is the difference between the x-coordinates of B and C. That would be -4 and -1, so the length of BC is [(-1)-(-4)] units = 3 units. With AB the situation is opposite. Since AB is a vertical line, we should find the difference between their y-coordinates instead. In this case, that would be (3-1) units = 2 units. Therefore, the area if triangle ABC is (3-2) sq. units = 1 sq. unit. Let's see what that is in mathematical expressions.

{\displaystyle {\begin{aligned}{\text{The length of }}\ BC&=x-(-z){\text{ units}}\\&=[(-1)-(-4)]{\text{ units}}\\&=3{\text{ units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The length of }}\ AB&=3-1{\text{ units}}\\&=2{\text{ units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The area of }}\ \triangle ABC&={\frac {BC\times AB}{2}}{\text{ sq. units}}\\&={\frac {3\times 2}{2}}{\text{ sq. units}}\\&=3{\text{ units}}\end{aligned}}}

Now we only have to figure out the area of the figure in the two bottom quadrants. The figure is shaped like a trapezium with a triangle stuck to its side and a hole in the middle. Using the knowledge we learnt previously, you should be able to figure out that a = -4 and b = -6. Note that we already know the value of -z from above, which is -2. Now we have to find the areas of the trapezium, the triangle, and the squared hole. Let's figure out the triangle first. Draw an imaginary line from Z perpendicular to XY, and name it R. The coordinates of R are (-4, -5) with reference to points X and Z. The lengths of RZ and XY should be 1 and 4 respectively, so the area of triangle XYZ is surely 2 sq. units. The details are presented below in mathematical expressions.

{\displaystyle {\begin{aligned}\because {\text{The }}x{\text{-coordinate of }}X=-4\\\therefore a=-4\end{aligned}}}

{\displaystyle {\begin{aligned}\because {\text{The }}y{\text{-coordinate of }}Y=-6\\\therefore b=-6\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The }}y{\text{-coordinate of }}D=-z\\=-2\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The length of }}\ ZR&=(-5)-(-4){\text{ units}}\\&=1{\text{ units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The length of }}\ XY&=(-2)-(-6){\text{ units}}\\&=4{\text{ units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The area of }}\ \triangle XYZ&={\frac {ZR\times XY}{2}}{\text{ sq. units}}\\&={\frac {1\times 4}{2}}{\text{ sq. units}}\\&=2{\text{ sq. units}}\end{aligned}}}

Now we have to find the area of the right-angled trapezium. Do you still remember the formula for the area of a trapezium? It is the sum of the parallel lines times the height, over two. We need to find the length of the parallel lines and height in order to find the area. We already know the height XY. The lengths of XD and YE are 9 and 8 respectively, so the area of the trapezium is 34 sq. units. The details are below.

{\displaystyle {\begin{aligned}{\text{The length of }}\ XD&=(-4)-5{\text{ units}}\\&=9{\text{ units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The length of }}\ YE&=(-4)-4{\text{ units}}\\&=4{\text{ units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The area of trapezium}}\ XYDE&={\frac {(XD+YE)\times XY}{2}}{\text{ sq. units}}\\&={\frac {(9+4)\times 4}{2}}{\text{ sq. units}}\\&=26{\text{ sq. units}}\end{aligned}}}

Our next step is to find the area of the square. The formula for the square is the length of one side squared, so the area of square FGHI is z2= 22 units = 4 units.

{\displaystyle {\begin{aligned}{\text{The area of square}}\ FGHI&=FI^{2}{\text{ sq. units}}\\&=z^{2}{\text{ sq. units}}\\&=2^{2}{\text{ sq. units}}\\&=4{\text{ sq. units}}\end{aligned}}}

The last part of point two is figuring out the total shaded area using the skills we learnt from previous chapters such as plane shapes. The first job is finding the total area of the figure in the third and fourth quadrants. The shaded area of the third and fourth quadrants = The area of pentagon XZYED - The area of square FGHI = The area of triangle XYZ + The area of trapezium XYDE - The area of square FGHI = (2+26-4) sq. units = 24 sq. units. Next, we add that to the area of triangle ABC, which gives us (24+3) sq. units = 11 sq. units. The mathematical expressions are shown below.

{\displaystyle {\begin{aligned}&{\text{The shaded area in quadrants III and IV }}\\&={\text{Area of }}\triangle XYZ{\text{ sq. units}}+{\text{Area of trapezium }}XZYED-{\text{Area of square }}FGHI\\&=(2+26-4){\text{ sq. units}}\\&=24{\text{ sq. units}}\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{The sum of all shaded areas }}&=(24+3){\text{ sq. units}}\\&=27{\text{ sq. units}}\end{aligned}}}

Since we have finished two goals already, it is time to find the third one. The third goal requires a technique that is the other way around the ones we used above. You may have noticed that the third goal is different from the others in that it requires the coordinates to be found, not a measure. Remember that the only coordinate we know from the question is those of point F. Also recall that all four sides of a square are the same. Since we can find the length of a segment given the coordinates of its endpoints, we can find the coordinates of the endpoints with the length of a segment.

So, how exactly do we do it? The coordinates of F are (-2, -3). We can find the y-coordinate of I which, we have learnt is the same as F's. Therefore, the y-coordinate is -3. But what about the x-coordinate? Since we know that FI = z = 2 units, we can work out that the x-coordinate of I is [(-2)+2)] units = 4 units. Remember that the position of the endpoint whose coordinate we are finding relative to the endpoint whose coordinate we already know is significant. If I were on the right of F, not on the left of it, its x-coordinate would be [(-2)-2)] units = -4 units. If I were above or below F, their x-cooridinates would be the same, and the y-coordinate of I would be [(-3)-2)] units = -5 units (below) or [(-3)+2)] units = -1 unit (above).

{\displaystyle {\begin{aligned}\because {\text{The }}y{\text{-coordinate of }}F=-3\\\therefore {\text{The }}y{\text{-coordinate of }}I=-3\end{aligned}}}

{\displaystyle {\begin{aligned}\because FI=z\\=2{\text{ units}}\end{aligned}}}
{\displaystyle {\begin{aligned}\therefore {\text{The }}x{\text{-coordinate of }}I=[(-2)+2)]\\&=4\end{aligned}}}

That's it! Not only have we learnt how to figure out the length and area of shapes and coordinates of points in a rectangular coordinate plane, but we have also seen an important example of how area is calculated.