平行線與垂直線的方程式 (英)

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平行線與垂直線的方程式 (英) : Equations of Parallel and Perpendicular Lines

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In this video we're going to do a couple of examples that
deal with parallel and perpendicular lines.
So you have parallel, you have perpendicular, and, of course,
you have lines that are neither parallel nor
perpendicular.
And just as a bit of review, if you've never seen this
before, parallel lines, they never intersect.
So let me draw some axes.
So if those are my coordinate axes right there, that's my
x-axis, that is my y-axis.
If this is a line that I'm drawing in magenta, a parallel
line might look something like this.
It's not the exact same line, but they have
the exact same slope.
If this moves a certain amount, if this change in y
over change in x is a certain amount, this change in y over
change in x is the same amount.
And that's why they never intersect.
So they have the same slope.
Parallel lines have the same slope.
Perpendicular lines, depending on how you want to view it,
they're kind of the opposite.
Let's say that this is some line.
A line that is perpendicular to that will not only
intersect the line, it will intersect it at a right angle,
at a 90 degree angle.
And I'm not going to prove it for you here.
I actually prove it in the linear algebra playlist. But a
perpendicular line's slope-- so let's say that this one
right here, let's say that yellow line has a slope of m.
Then this orange line, that's perpendicular to the yellow
line, is going to have a slope of negative 1 over m.
Their slopes are going to be the negative
inverse of each other.
Now, given this information, let's look at a bunch of lines
and figure out if they're parallel, if they're
perpendicular, or if they are neither.
And to do that, we just have to keep looking at the slopes.
So let's see, they say one line passes through the
points, 4, negative 3, and negative 8, 0.
Another line passes through the points, negative 1,
negative 1, and negative 2, 6.
So let's figure out the slopes of each of these lines.
So I'll first do this one in pink.
So this slope right here, so line 1, so I'll
call it slope 1.
Slope 1 is, let's just say it is-- well, I'll take this as
the finishing point.
So negative 3 minus 0-- remember change in y--
negative 3 minus 0, over 4 minus negative 8.
So this is equal to negative 3 over-- this is the same thing
as 4 plus 8-- negative 3 over 12, which is
equal to negative 1/4.
Divide the numerator and the denominator by 3.
That's this line.
That's the first line.
Now, what about the second line?
The slope for that second line is, well, let's take, here,
negative 1 minus 6, over negative 1 minus negative 2 is
equal to-- negative 1 minus 6 is negative 7, over negative 1
minus negative 2.
That's the same thing as negative 1 plus 2.
Well, that's just 1.
So the slope here is negative 7.
So here, their slopes are neither equal-- so they're not
parallel-- nor are they the negative
inverse of each other.
So this is neither.
This is neither parallel nor perpendicular.
So these two lines, they intersect, but they're not
going to intersect at a 90 degree angle.
Let's do a couple more of these.
So I have here, once again, one line passing through these
points, and then another line passing through these points.
So let's just look at their slopes.
So this one in the green, what's the slope?
The slope of the green one, I'll call that the first line.
We could say, let's see, change in y.
So we could do negative 2 minus 14, over-- I did
negative 2 first, so I'll do 1 first-- over 1
minus negative 3.
So negative 2 minus 14 is negative 16.
1 minus negative 3 is the same things as 1 plus 3.
That's over 4.
So this is negative 4.
Now, what's the slope of that second line right there?
So we have the slope of that second line.
Let's say 5 minus negative 3, that's our change in y, over
negative 2 minus 0.
So this is equal to 5 minus negative 3.
That's the same thing as 5 plus 3.
That's 8.
And then negative 2 minus 0 is negative 2.
So this is also equal to negative 4.
So these two lines are parallel.
They have the exact same slope.
And I encourage you to find the equations of both of these
lines and graph both of these lines, and verify for yourself
that they are indeed parallel.
Let's do this one.
Once again, this is just an exercise in finding slope.
So this first line has those points.
Let's figure out its slope.
The slope of this first line, one line passes
through these points.
So let's see, 3 minus negative 3, that's our change in y,
over 3 minus negative 6.
So this is the same thing as 3 plus 3, which is 6, over 3
plus 6, which is 9.
So this first line has a slope of 2/3.
What is the second line's slope?
This was the second line there, that's the other line
passing through these points.
So the other line's slope, let's see, we could say
negative 8 minus 4, over 2 minus negative 6.
So what is this equal to?
Negative 8 minus 4 is negative 12.
2 minus negative 6, that's the same thing as 2 plus 6.
The negatives cancel out.
So it's negative 12 over 8, which is the same thing if we
divide the numerator and the denominator by 4, that's
negative 3/2.
Notice, these guys are the negative
inverse of each other.
If I take negative 1 over 2/3, that is equal to negative 1
times 3/2, which is equal to negative 3/2.
These guys are the negative inverses of each other.
You swap the numerator and the denominator, make them
negative, and they become equal to each other.
So these two lines are perpendicular.
And I encourage you to find the equations-- I already got
the slopes for you-- but find the equations of both of these
lines, plot them, and verify for yourself that they are
perpendicular.
Let's do one more.
Find the equation of a line perpendicular to this line
that passes through the point 2 comma 8.
So this first piece of information, that it's
perpendicular to that line right over there, what does
that tell us?
Well, if it's perpendicular to this line, its slope has to be
the negative inverse of 2/5.
So its slope, the negative inverse of 2/5, the inverse of
2/5 is-- let me do it in a better color, a nicer green.
If this line's slope is negative 2/5, the equation of
the line we have to figure out that's perpendicular, its
slope is going to be the inverse.
So instead of 2/5, it's going to be 5/2.
Instead of being a negative, it's going to be a positive.
So this is a negative inverse of negative 2/5.
Right?
You take the negative sign and it becomes positive.
You swap the 5 and the 2, you get 5/2.
So that is going to have to be our slope.
And we can actually use the point-slope form right here.
It goes through this point right there.
So let's use point-slope form.
y minus this y-value, which has to be on the line, is
equal to our slope, 5/2 times x minus this x-value, the
x-value when y is equal to 8.
And this is the equation of the line in point-slope form.
If you want to put it in slope-intercept form, you can
just do a little bit of algebra, algebraic
manipulation. y minus 8 is equal to-- let's distribute
the 5/2-- so 5/2 x minus-- 5/2 times 2 is just-- 5.
And then add 8 to both sides.
You get y is equal to 5/2x.
Add 8 to negative 5, so plus 3.
And we are done.

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