34
THE SLOPE OF A
STRAIGHT LINE
Parallel and perpendicular lines
THE EQUATION OF A STRAIGHT LINE (Lesson 33) has this form --
y = ax + b
-- and its graph in general looks like this:
Now, the value of the y-coordinate depends on the value of the x-coordinate. Therefore as the x-coordinate changes, the y-coordinate will change. If the x-coordinate increases by 1, for example,
then in the line on the left, the y-coordinate increases much more than in the line on the right. We say that the line on the left has a greater slope than the line on the right. The slope of a straight line is a number that indicates the rate of change of the value of y with respect to the value of x. So many units of y for each unit of x.
If the x-axis represents time, and the y-axis distance, as is the case in many practical problems, then the rate of change of y with respect to x -- of distance with respect to time -- is called speed or velocity. So many miles per hour, or meters per second.
Thus, if the x-coordinate changes from the value x1 ("x sub 1") to the value x2 "x sub 2"), then we denote that change by the symbol Δx ("delta x"). Δx signifies the horizontal difference of the x-coordinates.
Δx = x2 − x1.
(Compare Lesson 32, The distance bewteen any two points.)
Similarly, Δy ("delta y") signifies the difference of the y-coordinates.
Δy = y2 − y1.
Δx is the horizontal leg of that right triangle; Δy is the vertical leg.
By the slope of a straight line, then, we mean this number:
| Slope | = | _Vertical leg_ Horizontal leg |
= | Δy Δx |
= |  |
For example:
If, when the value of x changes by 3 units, the value of y then changes
| by 2, then the slope of that line is | 2 3 |
. |
That means that for every 3 units that line moves to the right, it will move up 2. That is the rate of change between any two points on that line. Over 6 and up 4, over 15 and up 10. Because a straight line has one and only one slope. (Topic 8 of Precalculus.)
Up or down?
Which line do we say is sloping "up"? And which is sloping "down"?
Since we imagine moving along the x-axis from left to right, we say that the line on the left is sloping up, and the line on the right, down.
What is more, a line that slopes up has a positive slope. While a line that slopes down has a negative slope.
For, both the x- and y-coordinates of B are greater than the coordinates of A, so that both Δx and Δy are positive. Therefore their quotient, which is the slope, is positive.
But while the x-coordinate of D is greater than the x-coordinate of C, so that Δx is positive, the y-coordinate of D is less than the y-coordinate of C, so that Δy is negative. Therefore that quotient is negative.
Problem 1. What number is the slope of each line?
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| a) |  |
b) |  |
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| 3 4 |
4 3 |
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| c) |  |
d) |  |
|||||||
| − | 4 3 |
− | 3 4 |
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Horizontal and vertical lines
What number is the slope of a horizontal line -- that is, a line parallel to the x-axis? And what is the slope of a vertical line?
A horizontal line has slope 0, because even though the value of x changes, the value of y does not. Δy = 0.
| Δy Δx |
= | 0 Δx |
= | 0. | (Lesson 5) |
For a vertical line, however, the slope is not defined. The slope tells how the y-coordinate changes when the x-coordinate changes. But the x-coordinate does not change -- Δx = 0. A vertical line does not have a slope.
| Δy Δx |
= | Δy 0 |
= | No value. | (Lesson 5) |
Problem 2.
a) Which numbered lines have a positive slope? 2 and 4.
b) Which numbered lines have a negative slope? 1 and 3.
c) What slope has the horizontal line 5? 0.
c) What slope has the vertical line 6? It does not have a slope.
Example 1. Calculate the slope of the line that passes through the points (3, 6) and (1, 2)
Solution. To do this problem, here again is the definition of the slope. It is the number
| Δy Δx |
= | Difference of y-coordinates Difference of x-coordinates |
= | |
Therefore, the slope of the line passing through (3, 6) and (1, 2) is:
| Δy Δx |
= | 6 − 2 3 − 1 |
= | 4 2 |
= | 2 1 |
= 2. |
Note: It does not matter which point we call the first and which the second. But if we calculate Δy starting with (3, 6), then we must calculate Δx also starting with (3, 6).
As for the meaning of slope 2: On the straight line that joins those two points, for every 1 unit the value of x changes, the value of y will change by 2 units. That is the rate of change of y with respect to x. 2 for every 1.
Problem 3. Calculate the slope of the line that joins these points.
| a) | (1, 5) and (4, 17) | b) | (−3, 10) and (−2, 7) | |||||||||
| 17 − 5 4 − 1 |
= | 12 3 |
= | 4 | _ 10 − 7 _ −3 − (−2) |
= | 3 −1 |
= | −3 | |||
| c) | (1, −1) and (−7, −5) | d) | (2, −9) and (−2, −5) | |||||||||
| −5 − (−1) −7 − 1 |
= | −4 −8 |
= | ½ | −9 − (−5) 2 − (−2) |
= | −4 4 |
= | −1 | |||
The slope-intercept form
This linear form
y = ax + b
is called the slope-intercept form of the equation of a straight line. Because, as we can prove (Topic 9 of Precalculus): a is the slope of the line, and b is the y-intercept.
Problem 4. What number is the slope of each line, and what is the meaning of each slope?
a) y = 5x − 2
The slope is 5. This means that for every 1 unit x increases, y increases 5 units.
| b) y = − | 2 3 |
x + 4 |
| The slope is − | 2 3 |
. This means that for every 3 units x increases, |
y decreases 2 units.
Problem 5.
a) Write the equation of the straight line whose slope is 3 and whose
a) y-intercept is 1.
y = 3x + 1
b) Write the equation of the straight line whose slope is −1 and whose
a) y-intercept is −2.
y = −x − 2
| c) | Write the equation of the straight line whose slope is | 2 3 |
and which |
| passes through the origin. | |||
| y = | 2 3 |
x. The y-intercept b is 0. |
Problem 6. Sketch the graph of y = −2x.
This is a straight line of slope −2 -- over 1 and down 2 -- that passes through the origin: b = 0.
(Compare Lesson 33, Problem 15.)
The general form
This linear form
Ax + By + C = 0
where A, B, C are integers (Lesson 2), is called the general form of the equation of a straight line.
Problem 7. What number is the slope of each line, and what is the meaning of each slope?
a) x + y − 5 = 0
This line is in the general form. It is only when the line is in the slope-intercept form, y = ax + b, that the slope is a. Therefore, on solving for y: y = −x + 5. The slope therefore is −1. This means that for every 1 unit the value of x increases, the value of y decreases by 1.
| b) | 2x − 3y + 6 | = | 0 | |||
| −3y | = | −2x − 6 | ||||
| y | = | 2 3 |
x + 2, | on dividing every term by −3. | ||
| The slope therefore is | 2 3 |
. This means that for every 3 units |
the line goes over, it goes up 2.
| c) | Ax + By + C | = | 0 | |||||
| By | = | −Ax − C | ||||||
| y | = | − | A B |
x | − | C B |
, on dividing every term by B. | |
| The slope is − | A B |
. |
We can view that as a formula for the slope when the equation is in the general form. For example, if the equation is
4x − 5y + 2 = 0,
then the slope is
| − | 4 −5 |
= | 4 5 |
. |
Parallel and perpendicular lines
Straight lines will be parallel if they have the same slope. The following are equations of parallel lines:
y = 3x + 1 and y = 3x − 8.
They have the same slope 3.
Straight lines will be perpendicular if
| 1) | their slopes have opposite signs -- one positive and one negative, and | |
| 2) | they are reciprocals of one another. | |
That is:
| If m is the slope of one line, then a perpendicular line has slope − | 1 m |
. |
To be specific, if a line has slope 4, then every line that is perpendicular to it has slope −¼.
(We will prove that below.)
Problem 8. Which of these lines are parallel and which are perpendicular?
| a) y = 2x + 3 | b) y = −2x + 3 | c) y = ½x + 3 | d) y = 2x − 3 |
a) and d) are parallel. b) and c) are perpendicular.
Problem 9. If a line has slope 5, then what is the slope of a line that is
| perpendicular to it? | − |
1 5 |
| Problem 10. If a line has slope − | 2 3 |
, then what is the slope of a | |
| perpendicular line? |
3 2 |
||
| Problem 11. If a line has equation y = 6x − 5, then what is the slope | ||
| of a perpendicular line? | − |
1 6 |
The slopes of perpendicular lines
Theorem
If two straight lines are perpendicular to one another, then the product of their slopes is −1.
That is: If the slope of one line is m, then the slope of the perpendicular line
| is − | 1 m |
. |
Let L1 be a straight line, and let the perpendicular straight line L2 cross L1 at the point A.
Let L1 have slope m1, and let L2 have slope m2. Assume that m1 is positive. Then m2, as we will see, must be negative.
Draw a straight line AB of length 1 parallel to the x-axis, and draw BC at right angles to AB equal in length to m1.
Extend CB in a straight line to join L2 at D.
Now, since the straight line L2 has one slope m2 (Theorem 8.1 of Precalculus), the length of BD will be |m2|. For in going from A to D, we go over 1 and down |m2|.
That is, m2 is a negative number.
(Same figure.)
Angle CAD is a right angle. Therefore angle a is the complement of angle ß.
But triangle ABD is right-angled, and therefore the angle at D is also the complement of angle ß;
therefore the angle at D is equal to angle a.
The right triangles ABC, ABD therefore are similar (Topic 5 of Trigonometry),
and the sides opposite the equal angles are proportional:
This implies
m1 |m2| = 1.
But m2 is negative. Therefore,
m1m2 = −1.
Which is what we wanted to prove.
Next Lesson: Simultaneous linear equations
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