Subject: Geometry

# Coordinate Geometry

Coordinate Geometry involves the famous "Cartesian Coordinate Plane" by René Descartes which was developed on the 17th Century. It is a part of Geometry by which the points are located on the plane with the use of ordered pairs.

## What is a COORDINATE PLANE?

A coordinate plane consists of a horizontal axis (x-axis) and a vertical axis (y-axis) which intersects at a common point called origin, perpendicularly. The x and y axis is formed by a ''number line, by which the numbers on it corresponds to a point in the line which is then called coordinate of a point.

Figure 1: This is an example of a coordinate plane.
• All points or values that go to the right of the x-axis (after the origin) are always positive.
• All the points or values that go to the left of the y-axis (after the origin) are always negative.
• All points or values that go above the y-axis (after the origin) are always positive.
• All points or values that go down the y-axis (after the origin) are always negative.

### Ordered Pair

Ordered pairs are collectively called as "points" in the Coordinate plane. It consists of one coordinate point from the x-axis and one coordinate point from the y-axis, symbolized as (x,y). For example we have a point A with coordinates of (3,4) on the plane. Consider the picture below:

"3 is the x- coordinate, and 4 is the y-coordinate: collectively called as ordered pair"

- x- coordinate is also called Abscissa

-y-coordinate is also called Ordinate

## Plotting points in the coordinate plane

It is important in Coordinate Geometry that you will learn how to locate and plot points. Remember: X coordinate comes first (always) before y-coordinate in the ordered pair. Also, remember to name the point on the plot (if a point name is indicated on the given coordinates)

To plot points:

Suppose, we are given point B (4,2):

" (1) locate first the x-coordinate (the blue arrow), (2) then locate the y-coordinate (pink arrow), (3) after which, you then plot by putting a point on the area which represents (4,2)."

1. A ( 5,3)

2 B (3,2)

## Lines in Coordinate Geometry

lines in Coordinate geometry are formed when two or more points are connected. Suppose we have A (5,4) and B (2,1). When these points in the plane are connected it will form a line, as shown on the image below:

"When points A and B were connected, it formed line AB."

### Distance between Two points

The distance between two points is the length of the portion of the line/ line segment that connects one point to the other.

" the broken line is the distance between points X and Y."

### The Distance Formula

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

From the image above we have points: X (-5,4) and Y (3,3). Let us find the distance between these two points.

Let (x1, y1) be (-5,4), and (x2 , y2) be (3,3). Substitute the following values to the formula:

 d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} d = \sqrt{( 3-(-5) )^2+(3 - 4)^2} d = \sqrt{(8)^2+(-1)^2} d = \sqrt{64+1} d = \sqrt{65}

Thus, the distance between point X and point Y is \sqrt{65}

Let us have another example. Suppose we have points E ( -2, -5) and F (4,3), let us find the distance between point E and F. Let (x1, y1) be (-2, -5) and (x2 , y2) be (4,3). We then substitute the following values to the formula:

 d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} d = \sqrt{( 4-(-2) )^2+(3- (-5) )^2}$d = \sqrt{(6)^2+(8)^2}$ d = \sqrt{36 + 64}$d = \sqrt{100} d = {10} Thus, the distance between E and F is 10. ### Try to get the distance between 2 points on the following items: 1. A (-2, -3), B (2,4)  d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ d = \sqrt{( 2-(-2) )^2+(4- (-3) )^2}$d = \sqrt{(4)^2+(7)^2}$ d = \sqrt{16 + 49}$d = \sqrt{65}$

2. C (-3, -4), D (-2, 1)

 d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} d = \sqrt{( (-2) -(-3) )^2+(1 -(-4) )^2} d = \sqrt{(1)^2+(5)^2} d = \sqrt{1+25} d = \sqrt{26}

3. E (-5,5), F (5, -5)

 d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} d = \sqrt{( 5-(-5) )^2+( (-5) - 5 )^2} d = \sqrt{(10)^2+(-10)^2} d = \sqrt{100+100} d = \sqrt{200}

plot the following points, connect it (form a line) and get its distance:

4. G (-4,4). H (3, 5)

 d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} d = \sqrt{( 3-(-4) )^2+( 5 - 4 )^2} d = \sqrt{(7)^2+(1)^2} d = \sqrt{49+1} d = \sqrt{50}

5. I (-3,1), J (2, -4)

 d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} d = \sqrt{( 2-(-3) )^2+( (-4) - 1)^2} d = \sqrt{(5)^2+(-5)^2} d = \sqrt{25 + 25} d = \sqrt{50}

### Midpoint Formula

A midpoint is that which is halfway between two points.

In finding the coordinates of the midpoint of a line segment, we use the formula:

 x = 1\over{2} (x_1 +x_2) y = 1\over{2} (y_1 +y_2)

For example: Find the coordinates of the midpoint of A (5,4) and B (2,1).

Let (x1, y1) be (5,4), (x2, y2) be (2,1).

Substitute the following values to the formula.

 x = 1\over{2}(x_1 +x_2) x = 1\over{2}(5 + 2) x = 1\over{2}(7) x = 3.5
 y = 1\over{2}(y_1 +y_2) y = 1\over{2}(4 + 1) y = 1\over{2}(5) y = 2.5

Therefore, the coordinates of the midpoint of line AB is (3.5, 2.5)

Another example:

Find the coordinates of the midpoint of Y (-4,3) and Z (3, -1).

Let (x1, y1) be (-4,3), (x2, y2) be (3,-1).

Substitute the following values to the formula.

 x = 1\over{2}(x_1 +x_2) x = 1\over{2}((-4) + 3) x = 1\over{2}(-1) x = -0.5
 y = 1\over{2}(y_1 +y_2) y = 1\over{2}(3 + (-1)) y = 1\over{2}(2) y = 1

Therefore the coordinates of the midpoint of line YZ is (-0.5, 1)

### Try to answer the following:

Find the coordinates of the midpoint of:

1. (5,2), (3, -2)

Let (x1, y1) be (5,2), (x2, y2) be (3,-2).

Substitute the following values to the formula.

 x = 1\over{2}(x_1 +x_2) x = 1\over{2}(5 + 3) x = 1\over{2}(8) x = 4
 y = 1\over{2}(y_1 +y_2) y = 1\over{2}(2 + (-2)) y = 1\over{2}(0) y = 0

Thus, the coordinates are (4,0)

2. (2, 4), (3,0)

Let (x1, y1) be (2,4), (x2, y2) be (3,0).

Substitute the following values to the formula.

 x = 1\over{2}(x_1 +x_2) x = 1\over{2}(2 + 3) x = 1\over{2}(5) x = 2.5
 y = 1\over{2}(y_1 +y_2) y = 1\over{2}(4 + 0) y = 1\over{2}(4) y = 2

Thus, the coordinates are (4,0)

### SLOPE OF A LINE

The slope (m)of the line is the rise over the unit of run.

 rise = y_2 - y_1 run = x_2 - x_1

Therefore, the formula in getting the slope is:

 m = y_2 - y_1\over{x_2 - x_1}

For example:

Find the slope of the line that passes through points P (3,2) and Q (-5,-4). To solve, first we assign the values according to the formula. So, let (x1, y1) be (3,2), (x2, y2) be (-5, -4). Then, substitute the following values into the formula.

 m = y_2 - y_1\over{x_2 - x_1} m = (-4) - (-2)\over{(-5) - 3} m = (-2)\over{-8} m = 1\over{4}

Another example:

find the slope of the line that passes through points Q ( -5,3) and R ( 3, -2). Let, (x1, y1) be (-5, 3), (x2, y2) be (3, -2).

 m = y_2 - y_1\over{x_2 - x_1} m = (-2) - 3\over{3 - (-5)} m = (-5\over{8}

Slopes vary depending on to which the line rises.

* If a line rises to the right, then the slope is positive.

* If a line rises to the left, then the slope is negative

* If a line is horizontal then the slope is zero

* If a line is vertical its slope is undefined.

### Try to get the slopes of the following lines:

1. S (5,5), T (1,1)

 m = y_2 - y_1\over{x_2 - x_1} m = 1 - 5\over{1 - 5} m = (-4) \over{-4} m = 1

2. H (-4,5), I (3, -5)

 m = y_2 - y_1\over{x_2 - x_1} m = (-5) - 5\over{3 - (-4)} m = (-10)\over{-7} m = 10\over{7}

3. K (1,5) L (1,-2)

 m = y_2 - y_1\over{x_2 - x_1} m = (-2) - 5\over{1 - 1} m = (-7)\over{0} m = undefined

4. O (-5, -4), (3, -4)

 m = y_2 - y_1\over{x_2 - x_1} m = (-4) - (-4)\over{3 - (-5)} m = 0\over{8} m = 0