lines
lines
Vector and Parametric Equations of a Line
MCV4U: Calculus & Vectors
Recap
Determine vector and parametric equations of the line
parallel to ~r = (1, 4) + t(−2, 3) that passes through (3, −5).
The direction vector for ~r is (−2, 3).
Equations of Lines In Two-Space
Therefore, a vector equation of the new line is
r~2 = (3, −5) + s(−2, 3).
Part 2: Symmetric and Scalar Equations
The corresponding parametric equations are x = 3 − 2s and
y = −5 + 3s.
J. Garvin
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Symmetric Equation of a Line
Symmetric Equation of a Line
Parametric equations have the form x = xp + xm t and
y = yp + ym t, for some position vector ~p = (xp , yp ) are
~ = (xm , ym ).
direction vector m
Example
Rearranging each equation, t =
x−xp
xm
and t =
State the symmetric equation of the line defined by the
parametric equations x = −3 + 2t and y = 5 − 4t.
y −yp
ym .
x +3
y −5
and t =
, the equation is
2
−4
x +3
y −5
x +3
5−y
=−
or
=
.
2
4
2
4
Since t =
Since the parameter, t, is the same in both equations, we can
equate the right hand sides.
Symmetric Equation of a Line In Two Space
The symmetric equation of a line in two-space is
x−xp
y −yp
p = (xp , yp ) a position vector
xm = ym , where ~
~ = (xm , ym ) is a
representing a point on the line and m
direction vector for the line.
J. Garvin — Equations of Lines In Two-Space
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Symmetric Equation of a Line
Normal to a Line
Example
Another method of defining a line is to use a normal vector,
~n, which is perpendicular to the line.
State the symmetric equation of the line defined by the
equation ~r = (4, −1) + s(3, 7).
A position vector is ~p = (4, −1) and a direction vector is
x −4
y +1
~ = (3, 7), so the symmetric equation is
=
.
m
3
7
Recall that perpendicular lines in two-space have negative
reciprocal slopes.
Example
State a normal vector to ~r = (1, −2) + t(−3, 4).
~ = (−3, 4), ~n = (4, 3).
Since m
Using the dot product to verify, −3(4) + 4(3) = 0.
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Normal to a Line
Normal to a Line
Example
For slope-intercept form, solve for the y -intercept, b, in
y = mx + b.
Represent the equation of the line with normal vector
~n = (2, 5) that passes through P(−1, 3) using parametric,
vector, and slope-intercept forms.
3 = − 25 (−1) + b
b=
A direction vector for the line is perpendicular to ~n, so
~ = (−5, 2).
m
13
5
y = − 25 x +
13
5
The parametric equations of the line are x = −1 − 5t and
y = 3 + 2t.
The vector equation is ~r = (−1, 3) + t(−5, 2).
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Scalar Equation of a Line
Scalar Equation of a Line
Multiplying both sides of the equation by 5 and moving all
terms to one side, we obtain the equation 2x + 5y − 13 = 0.
Example
Determine the scalar equation of the line with normal vector
~n = (8, −3) that passes through P(−2, 1).
The equation, now written in scalar form (or standard form),
has the normal vector, ~n, as the coefficients of x and y .
Use A = 8 and B = −3 in the equation.
Scalar Equation of a Line In Two Space
8(−2) − 3(1) + C = 0
The scalar equation of a line in two-space is
Ax + By + C = 0, where ~n = (A, B) is a normal to the line.
C = 19
8x − 3y + 19 = 0
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Scalar Equation of a Line
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Questions?
Example
A line in two-space has scalar equation 2x − 5y − 15 = 0.
Determine the parametric equations of the line.
A normal to the line is ~n = (2, −5), so a direction vector for
~ = (5, 2).
the line is m
Substitute x = 0 (or any value) into the equation to find a
point on the line.
2(0) − 5y − 15 = 0
y = −3
Thus, parametric equations for the line are x = 5t and
y = −3 + 2t.
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