Algebra: Slope 2 Second part of determining the slope of a line
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- Hello.
- We are now gonna do some slope and y intercept problems as well.
- y-intercept problems as well.
- Let's get started.
- So let me make up a problem.
- Let's say we have the points two, five.
- The other point, let's make that negative three, negative three.
- Well, first let's just graph those two points.
- I'm going to graph them in yellow.
- So two, five.
- Let's see that's one two.
- One, two, three, four, five.
- So two, five is going to be right over there.
- OK.
- And then let me graph negative three, negative three.
- So it's one, two, three.
- One, two, three.
- So negative three, negative three is right over there.
- And then now let me draw a line that will connect them.
- That's my new technique.
- I draw it in two pieces.
- I think that's good enough.
- OK.
- So let's see if we can at least first figure out the slope of
- the line, and then if we have time we'll try to figure
- out the y-intercept.
- And then we'll know the whole equation for the line.
- Let me pick a slightly thinner color, and we'll get started.
- So the slope, if you saw the last module that just
- introduces how we calculate the slope, that's
- just rise over run.
- Or, change in y over change in x.
- This is y.
- So let's just do that real fast.
- So let's take this as our starting point.
- So change in y could be five-- remember, y is the
- second coordinate-- five minus negative three.
- And that's this one.
- Over-- now you do the change in x-- two minus, this
- is also negative three.
- Well five minus negative three, that's five plus plus three.
- So that equals eight.
- And then two minus negative three.
- Once again that's two plus plus three, so that equals five.
- So we figured out the slope of this equation.
- It's eight / five.
- And let's see if that makes sense.
- Let's figure out what the rise and the run is.
- If we were to start at this point right here, let's see how
- much we have to rise to get to the same y-coordinate
- as the other point.
- So let's see.
- We're here, and the other point is up here.
- So let's figure out what this distance is.
- Actually, now is a good time to use the fat.
- Oh man, I have a shaky hand.
- OK.
- Let's figure out what that distance is.
- That distance is delta y, which is change in y.
- So it's one, two, three, four, five, six, seven, eight.
- That equals eight.
- And that makes sense, because if you think about it
- what did we just do?
- We just took y equals five, which was up here, minus
- y equals negative three.
- And so obviously we just calculated that distance
- just by looking at the two coordinates five minus negative three.
- When you do this calculation it actually gives you
- this distance right here.
- So that's how we figure out how much we have to rise.
- So now let's do the run.
- Well the run, to go from this point to the other
- point, we went this far.
- And let's count how far that is.
- Well, it's one, two, three, four, five units.
- So we can say delta x is equal to five.
- And that's exactly what we did. delta y over delta x was equal
- to eight / five, or rise over run is equal to eight / five.
- And it would have been the same thing if we calculated run here
- or if we calculated rise here.
- But it's the same thing.
- Hope that's making sense to you.
- And I hope that also makes sense that if the rise for a
- given run becomes more, then the slope of the line is going
- to become steeper and it'll become a bigger number.
- So let's see what we have so far for the
- equation of this line.
- So so far we know the equation of this line is equal to, y is
- equal to the slope eight / five x plus b.
- So we're almost done.
- We just have to figure out this b right here.
- Now that b, just so you remember, that's
- the y-intercept.
- And that's where we intersect the y-axis.
- And since this graph is pretty neat, we can actually inspect
- it and see that, well, it looks like we're intersecting
- the y-axis at two.
- So my guess is we're going to come up with b equals two.
- But let's solve it, just in case we didn't have this
- neatly drawn graph here.
- So how can we solve for b?
- Well, we can substitute values that we know
- that work for x and y.
- Well either of these points are on that line, so we can
- substitute them in for x and y.
- So let's use the first one.
- OK.
- So the y we get five, will equal eight / five times x.
- Well, x there is two.
- Times two plus b.
- Well, now we just get five is equal to sixteen / five plus b.
- And then we get b equals-- well five is twenty-five / five, right?
- five is twenty-five / five minus sixteen / five equals nine / five.
- All right.
- See, so I was actually wrong.
- When I looked at this graph I said, oh that looks like
- almost two, so yeah it's probably going to be two.
- But when we actually did it using algebra, when we did it
- analytically, we actually saw that b is equal to nine / five.
- So it's almost two.
- 9/5 is 1 and 4/5, or 1.8.
- So that's almost two, but it actually turns
- out that it's not.
- It's at 1.8.
- And I can write it down as a decimal.
- 1.8.
- So the final equation for the line, I'm going to try to
- squeeze it in at the bottom of this page, it's going to be y
- is equal to-- well, we know the slope.
- eight / five x.
- Now we just add the y-intercept.
- Plus nine / five.
- There.
- We solved it.
- Let's do another one.
- And so-- that's nine / five.
- I don't want to be too repetitive.
- Let's do another problem.
- Time to do another problem, and let me put that
- graph back there again.
- There you go.
- All right.
- I'm going to think of two random numbers again.
- Let me try to do this fast, because YouTube puts a
- ten minute limit on me.
- So let's say I had the points two, negative three.
- And I had the point negative four, five.
- So two, negative three.
- Let's plot that sucker real fast.
- So x is two, so it's here.
- And the negative three.
- One, two, three.
- So two, negative three is there.
- And negative four, five.
- So that's one, two, three, four.
- One, two, three, four, five.
- I have to count like this because this
- graph is unlabeled.
- But if we actually were to draw in the coordinates you would
- that see this is five, and this is negative four, and so on.
- And this is two, and this is negative three.
- And now let's just draw a line.
- Let's draw it right there with my shaky hand.
- OK.
- There you go.
- Good line.
- And another good line.
- All right.
- So first we need to figure out the slope.
- Well we could just do that doing the algebra.
- So its slope is just delta-- I'm still using the line tool
- again-- delta y over delta x.
- Change in y over change in x.
- Let's take this y as the first point now.
- So we'll say five minus this y, negative three.
- Over-- now since we used the five first we have to use the
- negative four first as well.
- Negative four minus two.
- Well five minus negative three, that equals eight.
- And negative four minus two, well that equals negative six.
- And negative eight / six, well that equals-- they're
- both divisible by two.
- So that equals minus four / three.
- And let's see, does that make sense as the slope?
- Well, if we were to go down four from this point.
- So if the rise was negative four-- one, two, three, four.
- So if we go down-- woops, I'm using white.
- So that's why you can't see it.
- We go down by four here, and then we go to the right
- three, positive three.
- We still end up on the line.
- So it works.
- Looks good to me.
- Let's see if I can solve the y-intercept in thirty seconds.
- Otherwise, I'll start it on the next module.
- So we get y is equal to minus four / three x, plus b.
- And actually what we'll do is we'll leave off here, and I'm
- going to solve for b-- and you could try to do it on your
- own-- in the next installment of this presentation.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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