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)."

### Try to plot the following:

**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

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

Let (x_{1}, y_{1}) be (-5,4), and (x_{2} , y_{2}) 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 (x_{1}, y_{1}) be (-2, -5) and (x_{2} , y_{2}) 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 (x_{1}, y_{1}) be (5,4), (x_{2}, y_{2}) 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 (x_{1}, y_{1}) be (-4,3), (x_{2}, y_{2}) 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 (x_{1}, y_{1}) be (5,2), (x_{2}, y_{2}) 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 (x_{1}, y_{1}) be (2,4), (x_{2}, y_{2}) 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 (x_{1}, y_{1}) be (3,2), (x_{2}, y_{2}) 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, (x_{1}, y_{1}) be (-5, 3), (x_{2}, y_{2}) 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 |