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### Coordinate Geometry I

Lesson 18: Coordinate Geometry – Part I
Objectives

By the end of this lesson, students will be able to:

• locate points in the coordinate plane using their coordinates
• know and use the distance and midpoint formulas
• find the slope of a line
• find the slope of lines parallel and perpendicular to a given line
• prove theorems using coordinate geometry

Introduction

Applying algebraic principles to geometric figures is part of a branch of mathematics called coordinate geometry. A seventeenth-century French mathematician, Rene Descartes, is given credit for first developing this approach. In this lesson, we will review the coordinate plane and explore how it can help us understand more about geometry.

Coordinate Axes

You may have already done some of this work in your algebra course, but we will review some of this subject. We know that a number line consists of points that are associated with numbers, also called coordinates. Look at the number line below.

On the number line above, the point P has a coordinate of 0. We call this the origin. Point A has a coordinate of 2 and point B has a coordinate of -1. Since in geometry we study figures on a plane, we will set up a coordinate system so that every point on the plane has coordinates.

The coordinate plane is made by drawing two coordinate axes (number lines) perpendicular to each other. The point where the axes intersect is called the origin. The horizontal axis is called the x-axis and the vertical axis is called the y-axis. The coordinate axes divide the plane into four quadrants that are numbered as shown in the following figure.

Notice that the quadrants are numbered counterclockwise, starting in the upper right quadrant. The arrows in the picture above show the positive direction. The x-axis is positive to the right of zero and negative to the left. The y-axis is positive above zero and negative below zero.

We use the two axes to determine the location of a point on the plane. Every point on the plane has a pair of coordinates to show its location. The first number in the pair is the x coordinate. The second number is the y coordinate. They are written as (x, y).

For example, look at the coordinate plane below. Point P has coordinates of (4, 2). This means that point P is located four units to the right of zero and two units up from zero.

Look at the coordinate plane shown below. What are the coordinates of point A?

Since it is 5 units to the right of 0 and 3 units above 0, we describe the point as A (5, 3). You can also see that point A is in quadrant I.

How about points B, C, and D? Find their coordinates. In which quadrant is each point?

B (1, -4) is in quadrant IV.
C (-5, -2) is in quadrant III.
D (-2, 4) is in quadrant II.

Now look at point E. Its coordinates are (3, 0). Since it is on the axis, it is not in any quadrant. The same is true of point F, whose coordinates are (0, -2).

Practice

1 – 10. Using the figure below, find the coordinates of each of the following points.

1. A

A (7, 4)
From the origin, right 7 points, then up 4 points.

2. B

B (3, 9)
From the origin, right 3 points, then up 9 points.

3. C

C (0, 8)
From the origin, up 8 points on the y-axis.

4. D

D (-5, 5)
From the origin, left 5 points and up 5 points.

5. E

E (-7, 0)
From the origin, left 7 points on the x-axis.

6. F

F (-5, -4)
From the origin, left 5 points and down 4 points.

7. G

G (-2, -9)
From the origin, left 2 points and down 9 points.

8. H

H (0, -8)
From the origin, down 8 points on the y-axis.

9. I

I (3, -5)
From the origin, right 3 points and down 5 points.

10. J

J (3, 0)
From the origin, right 3 points on the x-axis.

Distance Between Two Points

When a football player throws a ball, how far is the football from the target? We can use the coordinate plane to answer questions like this one.

Look at the points A (3, 4) and B (5, 4). What is the distance between them?

The distance between the two points is 2 units. Since they both have the same y-coordinate, we simply subtract their x-coordinates.

5 – 3 = 2

We can say that line segment AB has a length of 2.

Now look at points C (-2, 6) and D (-2, 1).

The distance between the two points is 5 units. Since they both have the same x-coordinate, we subtract the y-coordinates.

6 – 1 = 5

We can say that line segment CD has a length of 5.

Since we are finding distance, the value cannot be negative. Always take the absolute value of the difference. This way, it does not matter which coordinate is subtracted from the other.

|6 - 1| = |5| = 5
|1 - 6| = |-5| = 5

In the two previous examples, either the x-coordinates or the y-coordinates were the same. Let’s see how to find the distance between any two points.

Look at the following graph. What is the distance between A (1, 4) and B (5, 1)?

Finding the distance between A and B is the same as finding the length of AB. To help find this distance, plot point C (1, 1). This forms right triangle ABC.

Now that we have a right triangle we can use the Pythagorean theorem to find the length of AB.

(AB)2 = (AC)2 + (CB)2
AC = 4 – 1 = 3
CB = 5 – 1 = 4
(AB)2 = 32 + 42 = 9 + 16 = 25
AB = 5

From this, we can derive a formula for finding the distance between two points. Let the first point P have the coordinates (x1, y1) and the second point Q have the coordinates (x2, y2).

The point R will have the coordinates (x1, y2). The length of QR is |x2 - x1|. The length of PR is |y2 - y1|. Let the distance from P to Q be called d.

d2 = |x2 - x1|2 + |y2 - y1|2
d2 = (x2 - x1)2 + (y2 - y1)2

 d = √ (x2 - x1)2 + (y2 - y1)2
Distance Theorem
The distance d between points (x1, y1) and (x2, y2) is equal to the following.
 d = √ (x2 - x1)2 + (y2 - y1)2

This is the theorem for the distance between two points. You can use this theorem for any two points on the coordinate plane.

For example, how would you find the distance between A (1, 5) and B (4, 9)?

Substitute the x and y values into the formula.

 d = √ (x2 - x1)2 + (y2 - y1)2
 d = √ (1 - 4)2 + (5 - 9)2
 d = √ (-3)2 + (-4)2
 d = √ 9 + 16 = √ 25 = 5

It does not matter which point you designate as (x1, y1) or (x2, y2). Look at this example again, this time with the coordinates switched.

d = √(1 - 4)2 + (5 - 9)2
d = √(-3)2 + (-4)2
d = √9 + 16 = √25 = 5

You can see that the answer is the same no matter which point is the starting point.

Example 1

Find the distance between A (4, 2) and B (-2, 10).

Substitute the values into the equation.

d = √(4 - (-2))2 + (2 - 10)2
d = √36 + 64 = √100 = 10

Example 2

Find the distance between A (5, 2) and B (8, 5).

Find the distance between A (5, 2) and B (8, 5).

d = √(5 - 8)2 + (2 - 5)2
d = √(-3)2 + (-3)2
d = √9 + 9 = √18 or 3 √2

Practice

1. (2, 6) and (12, 6)

(12 - 2)2 + (6 - 6)2  = 10

2. (-3, 0) and (-3, 5)

(-3 - (-3))2 + (5 - 0)2  = 5

3. (-5, 0) and (-8, 6)

(-8 - (-5))2 + (6 - 0)2  = √45 = 3 √5

4. (-4, -5) and (1, -2)

(-4 - 1)2 + (-5 - (-2))2  = √34

5. (-5, 1) and (7, 6)

(-5 - 7)2 + (1 - 6)2  = 13

Midpoint of a Line Segment

The midpoint is the exact center of a line segment.

To find the midpoint of a line segment, take the average of each coordinate.

Midpoint:
Given two points P (x1 + y1) and Q (x2 + y2), the midpoint of the line segment connecting the points is:
 ( x1 + x2 , y1 + y2 ) 2 2

For example, consider line segment AB. The coordinates of point A are (5, 6). The coordinates of point B are (3, 4).

To find the midpoint of AB, first find the average of the x-coordinates:

 5 + 3 = 4 2

Next, find the average of the y-coordinates:

 6 + 4 = 5 2

Therefore the midpoint of line segment AB is (4, 5).

Example 3

Find the midpoint of the line segment joining P (-2, 7) and Q (4, -3).

 -2 + 4 = 1 2
 7 + (-3) = 2 2

The midpoint is (1, 2).

Example 4

Find the midpoint of the line segment joining S (5, -1) and T (4, -2).

 5 + 4 = 4.5 2
 -1 + (-2) = -1.5 2

The midpoint is (4.5, -1.5).

Practice

1 – 5. Find the midpoint of the line segment connecting the following points.

1. (2, 6) and (8, 4)

 2 + 8 = 5 2
 6 + 4 = 5 2

The midpoint is (5, 5).

2. (2, 0) and (6, 9)

 2 + 6 = 4 2
 0 + 9 = 4.5 2

The midpoint is (4, 4.5).

3. (-4, 6) and (8, -4)

 -4 + 8 = 2 2
 6 + (-4) = 1 2

The midpoint is (2, 1).

4. (8, 7) and (6, -3)

 8 + 6 = 7 2
 7 + (-3) = 2 2

The midpoint is (7, 2).

5. (-2, -5) and (-7, -3)

 -2 + (-7) = -4.5 2
 -5 + (-3) = -4 2

The midpoint is (-4.5, -4).

Slope of a Line Segment

Have you ever been on a roller coaster, or at least seen one? The steepness of the hill is its slope. You learned about the slope of a line in algebra. Let’s review that concept in terms of coordinate geometry.

The most basic explanation of a slope is the phrase "rise over run." This phrase is not helpful for actually finding the slope, but it does clarify two things. One, it helps us think of a slope as a fraction. Two, for graphing, it helps us remember that the top number relates to the y-axis and the bottom number deals with the x-axis.

In coordinate geometry, when we talk about the slope of a line, we define it using any two points on the line.

Slope
The slope m of a line passing through two points, P (x1, y1) and Q (x2, y2) is given by:
 m = y2 - y1 x2 - x1
Where x2x1.

NOTE: Slopes should always be expressed as fractions in lowest terms, never as decimals. A slope expressed as a decimal is not helpful.

The numerator gives the change in the y-values between the two points. The denominator shows the change between the two x-values.

As with the midpoint formula above, it does not matter which point you designate as (x1, y1) or (x2, y2).

Find the slope of a line segment with endpoints of (4, 6) and (2, 2).

Insert the values into the formula.

 m = 6 - 2 = 4 = 2 4 - 2 2 1

Notice that you still reach the same answer if you substitute the values the other way.

 m = 2 - 6 = -4 = 2 2 - 4 -2 1

Example 5

What is the slope of the line segment with endpoints of (-2, 7) and (5, 1)?

Substitute the values and simplify.

 m = 7 - 1 = 6 = - 6 -2 - 5 -7 7

Compare the slope you found in Example 5 to the previous problem. In the previous problem, the line has a positive slope. When looking at it from left to right, the segment rises. Notice that m = 2, a positive number. In Example 5, the line falls as you look at it from left to right. This slope is negative.

Example 6

What is the slope of the line segment with endpoints of (3, 7) and (5, 7)?

 m = 7 - 7 = 0 = 0 3 - 5 -2

In this case, there is no change in the y-coordinate or the rise, so the slope is zero. All horizontal lines have a slope of zero.

Example 7

What is the slope of the line segment with endpoints of (-2, 4) and (-2, 1)?

 m = 4 - 1 = 3 -2 - (-2) 0

We cannot divide by zero. Therefore, we say that there is no slope, or that the slope is undefined. All vertical lines have an undefined slope.

To review, look at the four kinds of slope.

Practice

1 – 5. Find the slope of the line segments connecting each of the following pairs of points. State whether the slope is positive, negative, zero, or undefined.

1. (2, 6) and (8, 4)

 m = 4 - 6 = - 2 = - 1 8 - 2 6 3

negative

2. (2, 0) and (6, 9)

 m = 9 - 0 = 9 6 - 2 4

positive

3. (-4, 6) and (8, -4)

 m = -4 - 6 = -10 = - 5 8 - (-4) 12 6

negative

4. (8, 7) and (8, -3)

 m = -3 - 7 = - -10 8 - 8 0

undefined slope

5. (-2, -5) and (-7, -5)

 m = -5 - (-5) = 0 = 0 -7 - (-2) -5

zero slope

Parallel and Perpendicular Lines

When two lines are parallel, they will never intersect each other. Lines that are parallel have the same slope. Two lines that are perpendicular intersect each other at right angles. Their slopes are negative reciprocals of each other.

Consider the line that contains the points (1, 3) and (5, 6). Compare it to the line that contains the points (5, 1) and (9, 4). What is the relationship between the two lines?

First, find the slope of both. Then compare the values.

 m1 = 6 - 3 = 3 5 - 1 4
 m2 = 4 - 1 = 3 9 - 5 4

Since both lines have the same slope, the lines are parallel.

Example 8

Consider the line that contains the points (1, 4) and (3, -7). Compare it to the line that contains the points (5, 7) and (-6, 5). What is the relationship between the two lines?

Find the slope of the first line.

 4 - (-7) = - 11 1 - 3 2

Find the slop of the second line.

 7 - 5 = 2 5 - (-6) 11

Since the slopes are negative reciprocals of each other, the lines are perpendicular.

Practice

Find the slopes of the lines containing each of the following pairs of points. For each group, determine whether the lines are parallel, perpendicular, or neither.

1. (2, 2), (3, 1) and (5, -1), (7, -3)

 m1 = 2 - 1 = - 1 = -1 2 - 3 1
 m2 = -1 - (-3) = - 2 = -1 5 - 7 2

parallel

2. (-1, 3), (4, -7) and (1, 1), (-3, -1)

 m1 = 3 - (-7) = - 10 = -2 -1 - 4 -5
 m2 = 1 - (-1) = 2 = 1 1 - (-3) 4 2

perpendicular

3. (-3, 5), (4, 3) and (0, -2), (3, 6)

 m1 = 5 - 3 = - 2 -3 - 4 7
 m2 = -2 - 6 = -8 = 8 0 - 3 -3 3

neither

4. (9, -2), (-4, -2) and (-8, 3), (-8, 5)

 m1 = -2 - (-2) = 0 = 0 9 - (-4) 13
 m2 = 3 - 5 = -2 = no slope -8 - (-8) 0

perpendicular

5. (2, -5), (-4, 0) and (10, -6), (4, -1)

 m1 = -5 - 0 = - 5 2 - (-4) 6
 m2 = -6 - (-1) = - 5 10 - 4 6

parallel

Coordinate Proofs

Some theorems that we have derived are easier to prove using coordinate geometry. The key in using coordinate geometry is the placement of the figure on the coordinate axis. Keep the following in mind:

• Try to locate a key point at the origin. A key point could be a vertex or a midpoint of a segment.
• Try to locate a segment along one of the axes.
• Use as few letters as possible.

Look at the figure below.

To use this triangle ABC better for coordinate geometry, we will give the coordinates in terms of letters, such as A(a, b), B(c, d), and C(e, f).

Now we will relocate the triangle, keeping in mind the three points above.

By locating A at the origin, its coordinates will be (0, 0). Since B is on the x-axis, its coordinates can be (a, 0). C is located in quadrant I, so we will assign it the coordinates (b, c).

You can see that we reduced the number of letters used from 6 to 3 by relocating the triangle. Now we will find the coordinate of the midpoints of AC and BC.

 D, the midpoint of AC, has the coordinates ( b + 0 , c + 0 ) or ( b , c ). 2 2 2 2
 E, the midpoint of BC, has the coordinates ( a + b , c + 0 ) or ( a + b , c ). 2 2 2 2

Mark these points on the triangle.

Compare the length of DE to the length of AB

 DE = a + b - b = a 2 2 2

AB = a - 0 = a

In other words, the line segment that connects the midpoints of two sides of a triangle is one-half as long as the third side.

Now, find the slope of DE.

 m = c - c = = 0 2 2 0 a + b - b a 2 2 2

The slope of AB is also zero, since it is a horizontal line. This completes the proof of a theorem:

If a line segment connects the midpoint of two sides of a triangle, it is parallel to and half as long as the third side.

Try another proof using coordinate geometry. Can we prove that the diagonals of a parallelogram bisect each other? Look at the diagram below.

We will label point A (0, 0), point B (a, 0), and point D (b, c). Since B is “a” units from point A and ABCD is a parallelogram, we know point C is also “a” units from point D. Therefore, point C has the coordinates (b + a, c).

Find the midpoint of AC.

 ( b + a + 0 , c + 0 ) or ( b + a , c ) 2 2 2 2

Find the midpoint of DB

 ( b + a , c + 0 ) or ( b + a , c ) 2 2 2 2

Both line segments have the same midpoint. Therefore, they bisect each other. This gives us another theorem:

The diagonals of a parallelogram bisect each other.

Here is another proof by coordinate geometry. We will show that the midpoint of the hypotenuse is equidistant from the three vertices in a right triangle. In other words, using the diagram below, if D is the midpoint of CB, then CD = DB = DA.

First we will put our right triangle on the coordinate plane with the right angle at the origin.

Point A at the origin has the coordinates (0, 0). B is at (a, 0), and C has the coordinates (0, b). To find the coordinates of D, we use the midpoint formula:

 ( a + 0 , 0 + b ) or D ( a , b ) 2 2 2 2

Now we will find the length of AD, DC, and DB.

 AD = √ ( a - 0)2 + ( b - 0)2 2 2
 AD = √ ( a2 ) + ( b2 ) 4 4
 AD = 1 √ a2 + b2 2
 DB = √ ( a - a)2 + ( b - 0)2 2 2
 DB = √ ( - a )2 + ( b )2 2 2
 DB = √ ( a2 ) + ( b2 ) 4 4
 DB = 1 √a2 + b2 2
 DC = √ ( a - 0)2 + ( b - b)2 2 2
 DC = √ ( a )2 + (- b )2 2 2
 DC = √ ( a2 ) + ( b2 ) 4 4
 DC = 1 √a2 + b2 2

All the lengths are the same. We have verified that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices.

Now look at this problem. If you connect the midpoints of consecutive sides of a quadrilateral, what do you form?

We will label the points A (0, 0), B (a, 0), C (b, c), and D (d, e).

Now find the midpoints, E, F, G, and H.

 E = ( a + 0 , 0 + 0 ) = ( a , 0 ) 2 2 2
 F = ( a + b , 0 + c ) = ( a + b , c ) 2 2 2 2
 G = ( b + d , c + e ) 2 2
 H = ( d + 0 , e + 0 ) = ( d , e ) 2 2 2 2

Connect the midpoints. The figure with its coordinates is shown below.

Now, find the slope of EF, FG, GH, and HE.

 EF: m = c - 0 = c = 2 2 c a + b - a b b 2 2 2
 FG: m = c + e - c = e = 2 2 2 e b + d - a + b d - a d - a 2 2 2
 GH: m = c + e - e = c = 2 2 2 c b + d - d b b 2 2 2
 HE: m = e - 0 = e = 2 2 e d - a d - a d - a 2 2 2

What do you notice about the slopes? The slopes of EF and GH are the same. Also, the slopes of FG and GH are same. In other words, there are two pairs of parallel sides. A quadrilateral with two pairs of parallel sides is a parallelogram. So, what have we proven?

Connecting the midpoints of the consecutive sides of any four-sided figure will always form a parallelogram.

Practice

1. Prove by coordinate geometry that the diagonals of a rectangle are equal and that they bisect each other.

A(0, 0), B(a, 0), C(a, b), and D(0, b)
Find CA:d = √(a - 0)2 + (b - 0)2  = √a2 + b2
Find DB:d = √(a - 0)2 + (0 - b)2  = √a2 + b2
Therefore, the diagonals are equal in length.

 midpoint of CA = ( a + 0 , b + 0 ) = ( a , b ) 2 2 2 2
 midpoint of BD = ( a + 0 , 0 + b ) = ( a , b ) 2 2 2 2

Therefore, they have the same midpoint and bisect each other.

2. Prove by coordinate geometry that the median of a trapezoid is parallel to the base and one-half the sum of the parallel sides.

A(0, 0), B(a, 0), C(b, c), and D(d, c)
E is the midpoint of BC.

 E = ( b + a , c + 0 ) = ( b + a , c ) 2 2 2 2

F is the midpoint of DA.

 F = ( d + 0 , c + 0 ) = ( d , c ) 2 2 2 2

Find the slope of EF.

 m = c - c = = 0 2 2 0 b + a - d b + a - d 2 2 2

Therefore, EF is parallel to AB and DC.

 EF = b + a - d = b + a - d 2 2 2

AB = a - 0 = a
DC = b - d
AB + DC = a + b - d

 EF = a + b - d 2

Therefore, we know the length of EF.

 EF = 1 (AB + DC) 2

Enrichment Activity

To graph on a three-dimensional coordinate system, we need to add a third axis, the z-axis, to the two we already know. In drawing it, we let the y-axis and z-axis be on the plane. The x-axis is perpendicular to the plane. You can think about the x-axis as going into the screen and out towards you.

In plotting points on the 3-dimensional coordinate system, think about making a rectangular solid. We will put point A at the origin with coordinates A(0, 0, 0). Note that points have the coordinates (x, y, z). B is on the x-axis with coordinates B(4, 0, 0). D is on the y-axis with coordinates D(0, 3, 0).

We need to find the coordinates of C to form the base of our rectangular solid. Draw in lines parallel to the x- and y-axes. Where they intersect will be point C. The coordinates of C are C(4, 3, 0).

E is directly above the origin on the z-axis. E will have the coordinates E(0, 0, 5). H is directly above D with coordinates H(0, 3, 5). F and G are directly above B and C with coordinates F(4, 0, 5) and G(4, 3, 5).

Now use what you have just learned. Plot the point (2, 4, 3). Here is a hint: start with a point at the origin. Also, think about the position of the positive and negative parts of each axis.

Try graphing the following points.

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

Lesson Review

In this lesson, we covered the concept of coordinate geometry, taking algebraic principles and applying them to geometric figures. You learned the basic elements of coordinate geometry, like finding the distance between two points (d) and a line’s slope (m). Recall the formulas you used for this.

 d = √ (x2 - x1)2 + (y2 - y1)2
 m = y2 - y1 x2 - x1

You found the midpoint by averaging the coordinates. You also used the formulas to prove parallel lines (slopes are equal) and perpendicular lines (slopes are negative reciprocals). These were just some of the theorems you studied with coordinate geometry.

Homework

1 – 4. Use the figure below to answer the following questions.

1.

Which segment has a slope of zero?

2.

Which segment has no slope?

3.

Which segment has a positive slope?

4.

Which segment has a negative slope?

5 – 9. Use the points A (-3, -10) and B (5, 4) to answer the following questions.

5.

What is the slope of segment AB?

6.

What is the midpoint of segment AB?

7.

What is the length of segment AB?

8.

What is the slope of a line parallel to AB?

9.

What is the slope of a line perpendicular to AB?

10.

In the figure below, ABCD is a trapezoid. AB and CD are parallel.

1. What are the coordinates of C?
2. What is the midpoint of BD?
3. What is the midpoint of AC?
4. What is the length of AD?
5. What is the length of CB?