WEBVTT
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[MUSIC PLAYING]
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Hello.
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Welcome to the section of
exercises of the week three.
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Remember to try to solve
yourself the exercises
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before watching the videos, and
before looking at the solutions
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that you can find in the PDF
file at the bottom of the page.
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OK.
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It's Your Turn on the
geometry of the plane.
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We have the following exercise.
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We have to find the
equation of the line r
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passing through the point
of coordinates (6, 2)
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and forming with the y-axis an
angle theta equal to pi over 6.
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OK.
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Clearly our line is
not a vertical line,
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therefore its equation
has the following shape.
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y equal to m1 x plus q1.
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m1 is the slope.
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q1 is the y-intercept.
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But which is the meaning
of these two quantities?
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m1 is equal to the tangent of
the angle formed by our line r
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with the x-axis.
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That is the tangent
of this angle.
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But now, looking
at the picture, we
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have information
about the angle theta
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that our line forms
with the y-axis.
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And clearly, also, this
angle here is again theta.
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Therefore the angle we are
interested in between our line
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r and the x-axis
is equal to what?
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To pi over 2 minus theta.
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Then the slope m1 is
equal to the tangent
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of pi over 2 minus theta.
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That is, the tangent of
pi over 2 minus pi over 6,
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which is the tangent of what?
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We have six at the
denominator, and here we
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have 3 pi minus
pi, which is 2 pi.
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That is the tangent
of pi over 3.
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By definition, the
tangent of pi over 3
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is equal to the sine of pi over
3 over the cosine of pi over 3.
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The first is the square
root of 3 over 2,
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and the second, the
cosine, is 1 over 2.
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And we get the square root of 3.
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OK.
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This is m1.
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And what we can say about q1?
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q1 is the y-intercept.
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The y-intercept is
the second coordinate
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of the unique point belonging
to our line which belongs also
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to the y-axis.
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That is, q1, as you can read
easily from this equation,
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is exactly the value of
y when x is equal to 0.
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Good.
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Therefore let us
compute who is this q1.
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How can we compute q1 now.
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Substituting what?
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The coordinates of our point.
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You know (6, 2) are the
coordinates of a point
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belonging to our line.
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Therefore y equal to
2 and x equal to 6
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are solutions of this equation.
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And we immediately get 2 equal.
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OK. m1 we know is the
square root of 3 times
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the first coordinate, 6,
of our point, plus q1.
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And then we get that the
q1 is equal to 2 minus 6
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times the square root of 3.
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Good.
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And now we have our equation,
the equation of our line.
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The line r has equation y equal
to the square root of 3 times
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x, plus 2 minus 6 times
the square root of 3.
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OK.
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Then the second question is
to compute the intersection
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q of r with the y-axis.
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It's exactly the point we
were discussing before.
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It's that unique
point in the line r
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which has first
coordinate equal to 0.
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Then it's very easy, knowing
the equation of our line,
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to understand which are the
coordinates of this point q,
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because q is the point
which has first coordinate
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equals 0 and the
second coordinate
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equal to the y-intercept
of our equation--
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which is exactly 2 minus 6
times the square root of 3,
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as one can easily obtain
just setting x equals 0
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in the equation of the line.
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And then finally,
we want to compute
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the equation of the line s
passing through the point q
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and perpendicular to r.
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Therefore we want to compute
the equation of this line.
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It's clear that also the line
s is not a vertical line.
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Therefore, also the line s will
have an equation of the shape y
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equal to m2 x plus q2.
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OK.
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This time, we first
compute the y-intercept,
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because it's very easy.
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Because we know by the
question of the exercise
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that our line s has to
pass through the point q.
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Then q2 has to be equal to the
second coordinate of the point
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q, which is 2 minus 6
times the square root of 3.
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And what about the slope?
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When you have two
perpendicular lines--
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non-vertical lines,
both not vertical--
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then the slopes have to
satisfy a precise equation.
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Precisely, minus 1 has to
be equal to m1 times m2.
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OK.
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But we know that m1,
the slope or the line r,
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is equal to the
square root of 3.
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And then, from square root of
3 times m2 equal to minus 1,
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we get immediately
m2 equal to minus 1
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over the square root of 3.
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Then we can easily conclude that
the our line s has equation y
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equal to minus 1 over the
square root of 3 times x
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plus the y-intercept, which is
2 minus 6 times the square root
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of 3.
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This is the equation
of our line s.
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OK.
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Thank you very much
for your attention.
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