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www.mathcentre.ac.uk/.../the-gradient-of-a-straight-line-segment.mp4

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    The gradient of a line is
    a measure of how steep
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    that line is.
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    We may have a
    very steep line like
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    that.
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    And the gradient will
    be larger than a line.
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    Which is a bit more shallow.
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    So lines that are fairly shallow
    like this one will have fairly
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    Lopes, fairly small gradients,
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    steep lines. Have large
    gradients and lines.
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    Which are horizontal, will
    have zero gradients?
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    And we need to try to quantify
    that a little bit. Try and do
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    this a little bit more
    mathematically so we can
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    actually measure how much
    steeper this is than this one
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    than this one, and so on.
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    I'd run three line
    segments on this
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    diagram. Let's
    just look at them.
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    The first one.
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    Is the line segment from A to D?
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    Now as we move from A to D.
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    The X coordinate increases from
    one to two.
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    And the Y coordinate increases
    from one to five.
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    Now the line segment AD is
    steeper than the line segment
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    AC. As we move from A to C.
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    Exchange is from one to two and
    Y changes from one to three.
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    An AC in turn is steeper than
    the line segment AB. The line
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    segment AB is in fact
    horizontal because as X
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    increases from one to two, the
    Y coordinate doesn't change at
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    all. It remains at one.
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    Let's try and think about why
    mathematically, the line AD is
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    steeper than the line AC.
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    And the reason for this is that
    in both cases are X coordinate
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    is changing from one to two.
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    But as we move from A to D,
    there's a much bigger.
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    Change in Y than if we move from
    A to see so it's this relative
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    change in Y relative change in
    X, it's going to be important.
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    What we do is we calculate the
    change in Y.
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    Divide it by the
    change in X.
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    That's going to be a measure of
    the steepness. Let's do it for
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    the point AD for the points A&E.
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    OK, as we move from A to D.
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    Why changes from one to five? So
    the change in Y?
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    Is 5 - 1 and the change in X
    while exchanges from one to two.
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    So the change in X is 2 - 1.
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    So this quantity, the change
    in Y over change in X for the
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    line AD is 5 - 1, which is
    four 2 - 1 which is one and
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    four over one is 4.
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    So that's a measure of how much
    why changes as exchanges.
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    From one to two. What about
    the segment AC? Let's do
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    the same thing.
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    While the change in Y now is
    from one to three.
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    So the changes 3 - 1.
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    The change in X well X goes from
    one to two, so the changes 2 - 1
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    and again 3 - 1 is two. 2 - 1
    is one and 2 / 1 is 2. So this
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    is a measure of the relative
    change in X&Y.
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    What about a bee?
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    Well, as we move from A to B,
    why doesn't change at all? So
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    the change in Y is 1 - 1, which
    is of course 0.
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    And the change in X is still
    2 - 1, so we get zero over
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    one which is 0.
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    So you see this quantity change
    in Y divided by changing X gives
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    us a measure of the steepness of
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    these lines. As we would expect,
    the change in Y over change in X
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    for AD. Which we turned out to
    be 4 is greater than the change
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    in Y over change in X for AC
    because ady is steeper than AC
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    an intern. This change in Y over
    change in X for AC.
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    Is greater than the change in Y
    over change in X for a bee
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    because AC is steeper than a B?
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    So it's this quantity which
    gives us the measure that we're
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    looking for, and it's this
    quantity we define to be the
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    gradient of the line segment.
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    We often use the symbol M for
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    gradient. So the gradient is
    defined to be the change
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    in Y. Divided by
    the change in X.
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    As we move from one point to a
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    neighboring point. Let's do that
    for some general case.
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    Suppose we have.
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    System of coordinates
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    appoint a. X1Y
    One.
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    And a point B.
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    X2Y2 And
    we're interested in the gradient
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    of the line segment joining A&B.
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    Let me put in a horizontal line
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    through way. Anna vertical line.
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    Through be. So there's my X
    axis. There's my Y Axis.
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    As we move from A to B.
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    Exchange is from X one.
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    2X2.
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    Why changes from Y1?
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    To Y2.
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    So the change in Y divided
    by the change in X.
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    While the change in Y is the
    final value minus the initial
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    value, so it's Y 2 minus Y 1.
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    The change in X is X2
    minus X one.
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    And that is the formula that we
    can always use to find the
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    gradient of the line joining two
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    points. We can think of this
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    another way. Suppose we look at
    this angle in here. Let's call
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    that angle theater.
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    Now the change in Y is
    just this distance here.
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    Distance in there.
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    And the change in X is this
    distance in here.
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    And if we take the change in Y
    and divide it by the change in
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    X, what we actually get is the
    ratio of this side of this right
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    angle triangle to this side. And
    that's just the tangent of this
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    angle here. So this quantity
    that we've calculated is not
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    only the gradient of the
    line, it's also the tangent
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    of the angle that the line
    makes with the horizontal.
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    So the gradient M which we said
    is Y 2 minus Y, one over X2
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    minus X one is also equal to the
    Tangent Theta, where Theta is
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    the angle that the line makes
    with the horizontal.
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    We can take this a stage
    further. Suppose we continue
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    this line backwards until we
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    meet. The X axis.
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    And this angle in here between
    the extended line and the X
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    axis. Corresponds to this
    angle. Here these are
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    corresponding angles, so this
    two must also be theater.
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    So In other words, the gradient
    of the line is also the tangent
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    of the angle that the line makes
    with the X axis.
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    Let's have a
    couple of examples.
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    Let's choose a couple of points.
    Supposing a is the .34.
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    And B is the point.
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    814. Let's
    calculate the gradient of the
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    line joining these two points.
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    Well, the gradient is simply the
    difference in the Y coordinates
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    14 - 4.
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    Over the difference in the X
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    coordinates. 8 - 3.
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    14 - 4 is 10 and 8. Subtract
    3 is 5 and 5 into 10 goes
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    twice. So the gradient of this
    line is 2.
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    A second other example, suppose
    we have the point a, which
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    now has coordinates 04 and B
    which has coordinates 50. Let's
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    do the same calculation.
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    The gradient will be the
    difference in the Y coordinates.
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    That's 0 - 4.
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    Divided by the difference in the
    X coordinates 5 - 0.
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    So this time will get minus four
    on the top, five. At the bottom
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    we get minus four fifths.
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    So this is a little bit
    different now because we found
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    that we've got a negative number
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    for our gradient. And see
    what that actually means.
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    Let's plot the points and
    see what's going on.
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    Point A0X coordinate Y
    coordinate of four. So let's put
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    that there that's Point A.
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    Point B has an X coordinate of
    five that's there.
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    Y coordinate of 0, so there's my
    point there in there.
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    And the line joining them looks
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    like this. We know that
    this line has gradient
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    minus four fifths.
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    This line, as you notice, is
    sloping downwards as we move
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    from left to right.
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    And that's why the gradient
    turns out to be negative.
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    Another way of thinking about
    this is that the angle that the
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    line now makes with the X axis.
    This angle in here this theater
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    is now an obtuse angle.
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    Greater than 90 degrees less
    than 180 degrees, so we've an
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    obtuse angle. A line which is
    sloping downwards from left to
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    right. And a negative gradient.
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    Let me try to summarize
    all that behavior. If
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    you have a situation.
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    Like this?
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    Where the angle that the
    line makes with the
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    horizontal is acute.
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    Then the gradient.
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    Will be positive.
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    And the reason for that is that
    as you move along the line.
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    As X increases, why also
    increases so the change
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    in Y and the change in X
    have the same sign.
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    It's also important to recognize
    that if we take the tangent of
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    an acute angle, you get a
    positive number, so Tan Theater,
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    which we know is the same as
    them, is also positive.
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    What about an angle that
    sloping alignment sloping
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    downwards? We know that the
    angle is now.
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    Theater and it's obtuse.
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    We know that the tangent of an
    obtuse angle is negative, and as
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    we've seen, the gradient is
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    negative. And that's be'cause
    as X increases.
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    Why is decreasing so the change
    in Y and the change in X have
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    different signs, so we take the
    ratio, will find out that the
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    gradient is actually negative.
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    And finally.
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    Let's have one where theater is
    0, so the angle that the line
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    makes with the horizontal is 0,
    while tan feta is 0. So that's
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    consistent with our intuition.
    That tells us that MSO the
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    gradients 0. Let's
    have a look
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    at some parallel
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    lines.
    Here's
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    a
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    line.
    Let's call it L1.
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    L1 will make a certain angle.
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    Theater one with the X axis.
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    So it's gradient, as we've seen
    already, is Tampa Theatre 1.
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    M1, its gradient is tan.
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    Theater.
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    Let's put another line on this,
    also parallel to this first
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    line. This line is L2.
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    It will have a gradient M2.
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    That's extend it back to the.
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    Horizontal axis And let's
    measure this angle that would be
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    theater 2. And M2 will be the
    tangent of Theta 2.
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    Now, because these two
    lines are parallel.
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    They cross this X axis at the
    same angle, Theta one and three
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    to two. A corresponding angles
    Sophie to one must be equal to
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    three to two.
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    So because the to one is 3 to 2.
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    10 three to one what he called
    10 theater 2 so. In other words,
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    M1 equals M2.
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    So for two parallel lines,
    as you might have expected,
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    intuitively, the two
    gradients are equal.
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    And Conversely, if we have two
    lines for which the gradients
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    are equal. Then we can deduce
    from that that the two lines
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    must be parallel. OK, so that's
    parallel lines. Let's look at
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    some perpendicular lines.
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    See if we can do something
    about the gradients of
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    perpendicular lines.
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    Start with a point P.
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    And the origin there. And let's
    suppose point P has coordinates
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    a speed. That means that to get
    to pee from oh, we go a
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    distance a in the X direction
    and be in the Y direction.
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    Now what I'm going to do now is
    I'm going to draw a
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    perpendicular line, a line that
    is perpendicular to Opie, and
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    I'm going to do that by taking
    opian, rotating it through 90
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    degrees. So the point P will
    move around here and it will
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    move to appoint up there
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    somewhere. And let's call that
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    new point Q. I'm going to
    try to figure out what the
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    coordinates of QR.
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    This angle hit in here is 90
    degrees because Opie and oq are
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    perpendicular. Now to get from O
    to pee, wee had to go
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    horizontally a distance A and
    vertically be. So if this
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    triangle shifts around over here
    to get from Otak you will have
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    to go vertically a distance a.
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    And then horizontally a distance
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    be. So you see, we've just
    shifted this triangle, rotated
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    it round through 90 degrees, and
    doing that we can then read off
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    the coordinates of Point Q.
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    Q will have an X coordinate of
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    minus B. And AY
    coordinate of A.
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    That's now calculate the
    gradient of the line opi. Let's
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    call that MOPMLP. Remember, is
    the change in Y divided by
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    changing X as we move from OTP.
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    As we move from outer P, the
    change in Y is B minus zero.
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    The change in X is
    a minus zero.
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    So the gradient of Opie is
    just be over A.
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    What about the gradient of
    OQ? Let's call that MOQ.
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    We want the change in Y divided
    by the change in X as we move
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    from O to Q.
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    Well, the change in Y as we move
    from outer Q is a subtract 0.
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    And the change in X is minus
    B, subtract 0.
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    So this time this
    simplifies to a over minus
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    B or minus a over B.
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    Now let's see what happens when
    we multiply these two results
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    together. Let's take MOP and
    we're going to multiply it by
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    MCU. That's be over a
    multiplied by minus a over B.
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    And you see, when we do that,
    the aids cancel the beast
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    cancel, and we're left with
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    just. Minus one.
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    This is a very important result
    if you have two perpendicular
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    lines, then the product of their
    gradients is always minus one.
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    And correspondingly, if you've
    got 2 lines and you find that
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    when you multiply the gradients
    together, you get minus one, you
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    can deduce from that that the
    lines must be perpendicular.
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    Let's just have a look at an
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    example. Let's have three
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    points. Using A is the
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    .12. Because the
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    .34. And see is the point is
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    not 3. And we'll ask ourselves,
    the question is AB.
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    Perpendicular
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    To AC. Question is a
    be perpendicular to AC?
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    Well will will do this by
    calculating the gradient of the
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    line from A to B. Let's call
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    that MAB. And then we'll find
    the gradients of the line from A
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    to C will call that Mac.
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    So let's do this calculation. We
    want the gradient of the line
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    from A to B. Well, that's simply
    the difference in the Y
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    coordinates 4 - 2.
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    Over the difference in the X
    coordinates 3 - 1.
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    4 - 2 is two 3 - 1 is 2, so
    the gradient of a B is one.
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    What about the gradient from A
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    to C? Well, the difference in
    the Y coordinates now is 3 - 2.
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    The difference in the X
    coordinates is 0 - 1.
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    So we've got 3 - 2 is one
    0 - 1 is minus one.
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    So all this simplifies
    which is minus one.
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    If we multiply the two gradients
    together, maybe multiplied by
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    Mac, will get one times minus
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    one. Which is clearly minus one,
    so the gradients of these two
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    lines multiplied together have a
    result which is minus one, and
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    that means that the two lines a
    be an AC.
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    Indeed, must be perpendicular.
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www.mathcentre.ac.uk/.../the-gradient-of-a-straight-line-segment.mp4
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