1. Basic Definitions
For a right-angled triangle with hypotenuse length c,
We use this to find the distance between any two points (x1,y1) and (x2, y2) on the cartesian plane:
The point (x2, y1) is at the right angle. We can see that:
- The distance between the points (x1, y1) and (x2, y1) is simply x2 − x1 and
- The distance between the points (x2, y2) and (x2, y1) is simply y2 − y1.
Using Pythagoras' Theorem we have the distance between (x1, y1) and (x2, y2) given by:
Example 1:
Find the distance between the points (3, -4) and (5, 7).
Here, x1 = 3 and y1 = -4; x2 = 5 and y2 = 7
So the distance is given by:
Example 2:
Find the distance between the points (3, -1) and (-2, 5).
This time, x1 = 3 and y1 = -1; x2 = -2 and y2 = 5
So the distance is given by:
Gradient (or slope)
The gradient of a line is defined as
In this triangle, the gradient of the line is given by:
In general, for the line joining the points (x1, y1) and (x2, y2):
We see from the diagram above, that the gradient (usually written m) is given by:
Example:
Find the slope of the line joining the points (-4, -1) and (2, -5).
These are the points involved:
So the slope is:
Positive and Negative Slopes
In general, a positive slope indicates the value of the dependent variable increasesas we go left to right:
[The dependent variable in the above graph is the y-value.]
A negative slope means that the value of the dependent variable is decreasing as we go left to right:
Inclination
We have a line with slope m and the angle that the line makes with the x-axis is α.
From trigonometry, we recall that the tan of angle αis given by:
Now, since slope is also defined as opposite/adjacent, we have:
This gives us the result:
tan α = m
Then we can find angle α using
(That is, α = tan-1m)
This angle α is called the inclination of the line.
Example 1:
Find the inclination of the line with slope 2.
Here, tan α = 2, so
NOTE: The size of angle α is (by definition) only between 0° and 180°.
Example 2:
Find the slope of the line with inclination α = 137°.
The situation is as follows:
So the slope is:
m = tan α= tan 137°= -0.933
Note that the slope is negative.
Easy to understand math videos: MathTutorDVD.com
Let's see Gradient and Inclination using LiveMath.
Parallel Lines
Lines which have the same slope are parallel.
If a line has slope m1 and another line has slope m2 then the lines are parallel if
m1 = m2
Here is a LiveMath animation showing that if the gradient stays the same and we only change the y-intercept, the lines are parallel.
Perpendicular Lines
If a line has slope m1 and another line has slope m2then the lines are perpendicular if
m1 × m2= -1
In the example at right, the slopes of the lines are 2 and -0.5 and we have:
2 × -0.5 = -1
So the lines are perpendicular.
Example:
A line l has slope m = 4.
a) What is the slope of a line parallel to l?
b) What is the slope of a line perpendicular to l?
a) Since parallel lines have the same slope, the slope will be 4.
b) Using m1 × m2= -1, with m1 = 4, we obtain the value for m2:
Easy to understand math videos: MathTutorDVD.com
Special Cases
What if one of the lines is parallel to the y-axis?
For example, the line y = 3 is parallel to the x-axis and has slope 0. The line x = 3.6 is parallel to the y-axis and has an undefined slope.
The lines are clearly perpendicular, but we cannot find the product of their slopes. In such a case, we cannot draw a conclusion from the product of the slopes, but we can see immediately from the graph that the lines are perpendicular.
The same situation occurs with the x- and y-axes. They are perpendicular, but we cannot calculate the product of the 2 slopes, since the slope of the y-axis is undefined.
Exercises
(1) What is the distance between (-1, 3) and (-8, -4)?
(2) A line passes through (-3, 9) and (4, 4). Another line passes through (9, -1) and(4, -8). Are the lines parallel or perpendicular?
The line through (-3,9) and (4,4) has slope
The line through (9,-1) and (4,-8) has slope
Now
so the lines are perpendicular.
Note: We could have sketched the lines to determine whether they were parallel or perpendicular.
(3) Find k if the distance between (k,0) and (0, 2k) is 10 units.
This is the situation:
Applying the distance formula, we have:
Now so 5k2 = 100, giving:
k2 = 20
so
k = ± √20 ≈ ± 4.472
We obtained 2 solutions, so there are 2 possible outcomes, as follows:
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