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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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