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

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