Download 1 Parametric Lines

MA110
Eakin
Fall 2015
MA110: Fall 2015
Equations of Lines
Chapter 5 of text
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Parametric Lines
Suppose A(a1 , a2 ) and B(b1 , b2 ) are points in the plane E. Then, as when we think of points as complex
numbers we can add A and B and multiply them by real numbers. We agree to use the pair (a1 , a2 ) to
represent the point with coordinates (a1 , a2 ). Thus we also allow ourselves to write A = (a1 , a2 ) and
B = (b1 , b2 ).
A + B = (a1 , a2 ) + (b1 , b2 ) = (a1 + b1 , a2 + b2 )
λA = λ(a1 , a2 ) = (λa1 , λa2 )
Examples
(2, 4) + (−5, 7) = (2 − 5, 4 + 7) = (−3, 11)
7(−3, 5) = (−21, 35)
If z is a variable then we can multiply by z: zA = z(a1 , a2 ) = (za1 , za2 )
Definition: If L is a line the plane E then a linear parametric representation for L is a pair of
equations {x(t) = at + b, y(t) = ct + d} such that for each value of t the point (x(t), y(t)) is on L and
for each point P (r, s) which is on L then there is a value α of t such that r = x(α), s = y(α)
In this case {x(t), y(t)} is said to be a set of parametric representation for L.
Examples:
• {x(t) = t, y(t) = t} is a parametric representation for the line with equation y = x
Explanation: Pick a value of t, say t = 5. Then (x(5), y(5)) = (5, 5) is on the line y = x.
Conversely, if we pick a point on the line y = x, say (−4, −4) then this corresponds to t = −4.
• {x(t) = 2t − 1, y(t) = 2t − 1} is a different paametric representation for the line with equation
y = x.
Explanation: Pick a value of t say t = 5 then (x(5), y(5)) = (2 ∗ 5 − 1, 2 ∗ 5 − 1) = (9, 9) which
is a point on the line y = x.
Conversely, if we pick a point on the line y = x, say (−4, −4) then we can solve for the corresponding value of t. We need 2t − 1 = −4 so t = − 23 .
Obviously there are infinitely many linear parameterizations of the line y = x. The same is true
for any line. Moreover, there are non-linear parametrzations too. For instance the equations {x(t) =
t3 + 1, y(t) = t3 + 1} also provide a parametric description of the line y = x. For now we will stick to
linear parametric equations.
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Parmetric descriptions of the line through two points
If A and B are two points in the plane and t is a variabe then
P (t) = t B + (1 − t) A
This very useful compact form doesn’t immediately display the parametric equations but they drop
out quickly.
If A = (3, 7) and B = (2, −5) then
P (t)tA + (1 − t)B = t(2, −5) + (1 − t)(3, 7) =
(2t + 3(1 − t), −5t + (1 − t)7) = (3 − t, 7 − 12t)
So the x−coordinate of the point corresponding to the parameter t is 3 − t and the y−coordinate
os 7 − 12t. So the parametric equations are {x = 3 − t, y = 7 − 12t}.
2.1
Recovering the conventional “cartesian” equation for the line
In the previous example we found that {x = 3 − t, y = 7 − 12t} is a set of parametric equations for the
line through the points A = (3, 7) and B = (2, −5).
The value of t is the same in each coordinate and we solve each for t to get
x−3
= t = y−7
−1
−12
y−7
So x−3
=
−1
−12
(−12)(x − 3) = (−1)(y − 7) which we solve for y to get y = 12x − 29
check: We check that the coordinates of each of the points A and B satisfy this eq equation to
verify that it is correct:
For the point A(3, 7): 7 = 12 ∗ 3 − 29 is true and for B(2, −5): −5 = 12 ∗ (2) − 29 is true.
Remember that parametric representations are not unique. They carry with them the sense of motion along the line. If you think of the road from Lexington to Louisville as a straight line and you drive
it, keeping track of your position at every time t then you are producing a parametric representation
of the line.
The fact that our parametric equations are linear corresponds to the motion having constant velocity.
In the parametric description P (t) = tB + (1 − t)A we see that P (0) = A and P (1) = B.
In the equally valid parametrization Q(t) = tA + (1 − t)B we see that Q(0) = B and Q(1) = A
So P captures the idea of moving at constant velocity along a straight line from A to B in one unit of
time while Q corresponds to moving at constant velocity along the same straight line from B to A in
one unit of time.
Of course non-linear equations correspond to non-linear motion.
Remark: Parametric representations exist for all lines, including vertical lines.
If A = (1, 2) and B = (1, 3) are on a line then the line is vertical and has no point-slope equation
of the form y = mx + b. However the parametric form P (t) = tB + (1 − t)A
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2.2
Producing a parametric representation from a cartesian equation for
a line
If y = 2x + 3 is the point-slope cartesian equation for the line L then {x(t) = t, y(t) = 2t + 3} is a set
of parametric equations for L.
If L is vertical with equation x = 7 then {x(t) = 7, y(t) = t} is a set of parametric equations for L.
2.3
Parametric forms describe line segments
In many applications one wants to restrict attention to the points on a line segment AB rather than
the entire line L which contains it.
If P (t) = tB + (1 − t)A is the parametrization of L then the points between A and B are simply
the points corresponding to the values of the parameter between 0 and 1.
For 0 ≤ α ≤ 1, the point P (α) is the point on the line segment AB such that the distance from A
to P (α) is α times the distance from A to B.
When we write the line as P (t) = tB + (1 − t)A and observe that A corresponds to 0 then thing of
moving from from A to B at a constant velocity over one unit of time we will have gone, for instance,
1
of the way when t = 31 and 12 of the way when t = 12
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• P ( 12 ) = 21 A + 21 B is the midpoint of the line segment AB
The midpoint of the line segment connecting A = (3, 7) and B = (2, −5) is P ( 12 ) = 12 (3, 7) +
1
(2, −5) = ( 52 , 1)
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• P ( 31 ) = 31 A + 32 B is the point on the line segment AB that is
1
3
the length of AB from A
The points that divide the line segment connecting A = (3, 7) and B = (2, −5) into three equal
segments are 13 A + 32 B = ( 73 , −1) and 32 A + 13 B = ( 83 , 3)
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Alternate form for parametric equations and direction of a
line
Let L be the line through A and B with A 6= B.
The parameterization P (t) = tB + (1 − t)A can be re-written as
P (t) = A + t(B − A)
In this form it resembles the y = mx + b form with B − A playing the form of the slope and A the
intercept (the value at t = 0). We refer to B − A as a “direction” of L.
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• L has infinitely many directions as C − D is also a direction for any points C, D on L.
Conversely, if R(v) = W + vZ is a parameteriation of L with direction Z then W = R(1) − R(0)
so Z is the difference of two points on L. That is,
directions of L = {C − D | C and D are on Land, C 6= D}
• if r is any non-zero number and S(t) = G + tH is a parameterization of L. Then M (t)S(rt) =
G + (rt)H = G + t(rH) is also a parameterization and we see that if H is a direction for L then
rH is also a direction for L for any r 6= 0.
• If Z is any direction for L and C is any point on L then N (α) = C + αZ is a parametrization for
L.
Since Z is a direction for L then there is a parametrization of L of the form R(v) = W + vZ.
Since C is on L there is vo such that C = R(vo ) = W + vo Z so W = C − vo Z and R(v) =
C − vo Z + vZ = C + (v − vo )Z so if we let α = v − vo then K(α) = C + αW is a parameterization
of L.
• If L is a line and A(a1 , a2 ) and B(b1 , b2 ) are points on L then the direction B−A = (b1 −a1 , b2 −a2 ).
If a1 6= b1 then
b 2 − a2
m=
b 1 − a1
is the slope of the line L. Since a line has only one slope this means that if H = (h1 , h2 ) is any
direction of L then m = hh21 is the slope of L
• If L is a line and (a, b) and (c, d) are directions for L then there is a number λ such that
λ(a, b) = (c, d). This is often expresses as:
If L is a line in the plane then any two directions for L are proportional.
To see this we need a λ such that λa = c and λb = d
If there is such a λ and a, b 6= 0 then ac = λ = db . That is λ has to be
If a, b 6= 0 then
b
a
and
d
c
are both the slope of L so
b
a
=
d
c
c
a
(and db ) if these are equal.
and we are done.
If a L does not have a slope then neither ab or dc can be numbers so both a and c must be 0 so
(a, b) = (0, b) and (c, d) = (0, d) . Since (0, 0) can’t be a direction, b 6= 0 so we take λ = ad and
λ(0, a) = (0, d).
3.1
examples
• Find a direction W for the line L through A(1, 4) and B(7, −3) which is of the form (1, a) or
(a, 1).
solution B − A = (7, 3) − (1, 4) = (6, −1) is a direction for L. Any non-zero multiple of a
direction for L is also a direction so 61 (6, −1) = (1, − 16 ) and (−1) (6, −1) = (−6, 1) both work.
• Suppose (3, 5) is a direction for the line L which contains the point (4, 7). What is the point-slope
equation for L.
solution Since (4, 7) is a direction for L, 74 is the slope of L so the equation is y − 7 = 47 (x − 4)
•
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Geometric interpretations of the directions of a line
Parallel Lines
Two lines are parallel if and only if their directions are proportional
Of course lines have infinitely many directions so what this means is that any direction of one is
proportional to any direction of the other.
Let (a, b) and (c, d) be directions for L and M with (a, b) = λ(c, d) and λ 6= 0.
If either c or a is 0 then so is the other in which case the lines have no slope, are therefore vertical
λc
and are parallel. If neither a nor c is zero then from a = λc, b = λd we have ab = λd
= dc which says
that the lines have the same slope and are therefore parallel. Conversely, if two lines are parallel then
they are both vertical or they have the same slope. If they are vertical then their directions are of
the form (0, u) and (0, v) which are proportional. If they are not verical then each has a slope and
the two are equal since they are parallel. Call the common slope m. Then ab = m = dc so ad = bc so
c
(a, b) = (c, cba ) = (c, ad
) = (c, d)
a
a
A direction can be interpreted as a non-zero point in the plane. With this view
If L is a line in the plane then the directions of L are all of the non-zero points on the
line through the origin that is parallel to L
Perpendicular Lines
If (a, b) is a direction for a line L then (a, b)⊥ = (−b, a) is a direction for any line perpendicular to L.
For non-vertical lines, non-horizontal lines a, b 6= 0. Then m =
b
a
is the slope of L and
a
−b
= − 1b =
a
− m1 is the negative reciprocal of m and hence the slope of a line perpendicular to L.
If (a, b) = (0, b) then L is a vertical line and (a, b)⊥ = (−b, 0) is the direction of a horizontal line which
is then perpendicular to L.
There is a simple way to use directions to tell if two lines are parellel.
If (a, b) and (c, d) are directions for lines L and M then L is perpendicular to M if and
only if ad + bc = 0
subsectionExamples
• Problem: Find a parametric description for the line N through C(−7, 2) that is parallel to the
line L passing through A(3, 4) and B(5, −9)
Solution: A direction for L is B − A = (5, −9) − (3, 4) = (2, −13).
The line through C with this direction is P (t) = C + t(B − A) = (−7, 2) + t(2, −13)
• Problem: Find a parametric description for the line M through C(−7, 2) that is perpendicular
to the line L passing through A(3, 4) and B(5, −9)
Solution: A direction L is B − A = (5, −9) − (3, 4) = (2, −13).
So a direction for a perpendicular line is (B − A)⊥ = (13, 2).
So Q(s) = C + s(B − A)p erp = (−7, 2) + t(13, 2) is a parametric form of the line M
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• Problem Find the parametric form of W , the perpendicular bisector of the line segment with
end points A(2, 4) and B(5, 9).
Solution: A direction for the line containing this segment is B − A = (3, 5) so (3, 5)⊥ = (−5, 3)
is a direction for any perpendicular line.
) and W must
The midpoint of the line segment is M = 21 A + 12 B = 21 (2, 4) + 12 (5, 9) = ( 72 , 13
2
contain it so a parametric from for W is
7 13
( , ) + t(−5, 3)
2 2
Which gives paametric equations x(t) =
7
2
+ 5t, y(t) =
13
3
+ 3t
• Problem: Find the point of intersection of the line L which passes through A(2, −3) and B(4, 7)
and the line M which passes through C(6, 5) and D(−8, 1).
Solution A parametric form for L is A + t(B − A) = (−2, 3) + t(6, 4) = (−2 + 6t, 3 + 4t)
A parametric form for M is C + s(D − C) = (6, 5) + s(−14, 4) = (6 − 14s, 5 − 4s)
IMPORTANT: Note that we have used different parameters for M and L. This is important!
The problem asks us to find a point common to L and M . If we think of the parametric forms
as the paths traversed by Sally as she walks along M and Joe as he walks along L then different
parameters, t and s corresponds to asking if they passed the same point on their separate walks.
Using the same parameter, say t, corrsesponds to asking if they were at the same point at the
same time.
Solution: The parametric equations are
L:
x = −2 + 6t, y = 3 + 4t
M:
x = 6 − 14s, y = 5 − 4s
At the point of intersection x = x and y = y so
−2 + 6t = 6 − 14s
3 + 4t = 5 − 4s
Solving these two linear equations in two variables we get s = 85 , t = −1
8
Setting s = 85 in the equations for M we get x = −11
, y = 52
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As a check we set t = −1
in the equations for L and get the same thing. Thus the point of
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5
intersection is ( −11
,
).
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2
5
Reflections
If a particle travels in the plane along a line L which has direction D and it meets a line M which
has direction A then it reflects off A and travels along a line R. It is easy to calculate the point of
intersection of L and M so if we have the direction of R we can calculate the reflected path of the
particle. In fact this is very easy to do.
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• The direction D can be written in the form D = αA + βA⊥
We will see how to do this below. The idea is that the motion along L can can be regarded
as being made up of two components: one parallel to M and one perpendicular to M . The
component parallel to M ( the αA part) isn’t changed so it remains the M component of the
reflection. The component perpendicular to M , (βA⊥ is reflected back in the opposite direction
(−βA⊥ ) so that is the A⊥ component of the reflection. Thus we have:
• The direction of R is αA − βA⊥
5.1
Example
Problem: The line L passing through A(3, 7) and B(9, −5) meets the line M which passes through
C(1, 3) and D(4, 8) and is reflected. Find parametric equations for the line of reflection.
Solution: A direction of L is B − A = (6, −12) and a direction of M is D − C = (3, 5)
We want to resolve the direction of L into components parallel and perpendicular to M so we write
(6, −12) = α(3, 5) + β(3, 5)⊥ = α(3, 5) + β(−5, 3)
(6, −12) = (3α − 5β, 5α + 3β)
3α − 5β = 6
5α + 3β = −12
Solving these we get β =
−33
,
17
α = − 21
17
so (6, −12) = − 21
(3, 5) + −33
(−5, 3)
17
17
and the direction of the reflection is
(3, 5) − −33
(−5, 3) = ( −288
, −6
)
− 21
17
17
17
17
So now we need the point of intersection.
Parametric equations for the two lines are:
L : (3, 7) + t(6, −12) = (3 + 6t, 7 − 12t) so x(t) = 3 + 6t, y(t) = 7 − 12t
M : (1, 3) + s(3, 5) = (1 + 3s, 3 + 5s) so x(s) = 1 + 3s, y(s) = 3 + 5s so
x : 3 + 6t = 1 + 3s
y : 7 − 12t = 3 + 5s
Solving these two linear equations in two unknowns (e.g. use Cramer’s Rule) we get t =
Using the equation for M we get that the point of intersection is
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(1, 3) + 11
(3, 5) = ( 35
, 73 )
11 11
Thus the line of reflection is ( 35
, 73 ) + t( −288
, −6
)
11 11
17
17
35
−288
or x(t) = 11 + t 17
+ t −6
y(t) = 73
11
17
If we solve each of these for t, equate the two values of t and simplify we get
1
y = 48
x + 3469
which is the cartesian equation for the line of reflection.
528
7
1
,s
33
=
73
11
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Lines in higher dimensions
Since lines have equations of the form y = mx + b in the plane, one is led to ask about equations for
lines in three dimensions. There an equation of the form z = ax + by describes a plane rather than a
line. For instance the equation z = 0 describes the x − y plane. Indeed there is no way to describe
a line in three dimensions with a single cartesian equation. That is why the parametric form has an
enormous advantage over the cartesian form as soon as one leaves the plane.
In three dimensions a parametric form of the line through A and B is
P (t) = A + t(B − A)
which looks exactly like the form in the plane. The only difference is that directions, like points, have
three coordinates rather than two. In fact this works in any number of dimensions.
Example Problem: What is the parametric form of the line in three dimensions through A(2, 3, 4)
and B(7, −9, 3) ? Solution: P (t) = A + t(B − A) = (2, 3, 4) + t((7, −9, 3) − (2, 3, 4)) = (2, 3, 4) +
t(5, −12, −1)
This gives parametric equations (x(t), y(t), z(t)) = (2 + 5t, 3 − 12t, 4 − t) or
x(t) = 2 + 5t,
y(t) = 3 − 12t,
z(t) = 4 − t
As in the plane, parallel lines are lines that have the same directions.
The situation for perpendicular lines is more complicated in higher dimensions since at any point
there is an entire plane that is perpendicular to a given line (e.g. the x − y plane is prrpendicular to
the z− axis).
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