Calculus and Vectors – How to get an A+
8.2 Cartesian Equation of a Line
A Symmetric Equation
The parametric equations of a line in R2:
⎧ x = x0 + tu x
t∈R
⎨
⎩ y = y0 + tu y
may be written as:
x − x0 y − y 0
=
= t, t ∈ R
ux
uy
The symmetric equation of the line is (if exists):
x − x0 y − y 0
=
ux
uy
Note. The symmetric equation does exist if
u x ≠ 0 and u y ≠ 0 .
Ex 1. Convert the vector or parametric equations to the
symmetric equation (if exists).
⎧ x = −3 + 2t
a) ⎨
t∈R
⎩ y = −5t
x+3 y−0
=
= t, t ∈ R
2
−5
x+3
y
∴
=
2
−5
r
b) r = (0,1) + t (1,−2), t ∈ R
⎧x = t
x y −1
t∈R ⇒ ∴ =
⎨
1 −2
⎩ y = 1 − 2t
r
c) r = (−2,3) + t (2,0), t ∈ R
⎧ x = −2 + 2t
x+2 y −3
t∈R ⇒
=
⇒∴ does not exist
⎨
y
=
3
2
0
⎩
Ex 2. Convert the symmetric equation of a line L :
x + 2 y −3
=
to the vector equation.
−1
2
x + 2 y −3
=
= t, t ∈ R
−1
2
⎧ x + 2 = −t ⎧ x = −2 − t
r
⇒⎨
⇒ ∴ r = (−2,3) + t (−1,2), t ∈ R
⎨
⎩ y − 3 = 2t
⎩ y = 3 + 2t
B Normal Equation
Let consider a line L that passes through the
specific point P0 ( x0 , y 0 ) and has the direction
r
vector u = (u x , u y ) .
C Cartesian Equation
The normal equation can be written as:
r r r r
r ⋅ n − r0 ⋅ n = 0
( x, y ) ⋅ ( A, B) − ( x0 , y 0 ) ⋅ ( A, B) = 0
Ax + By − Ax0 − By0 = 0
Ax + By + C = 0 where C = − Ax0 − By0
The Cartesian equation of a line is given by:
Ax + By + C = 0
where
r
n = ( A, B) is a normal vector and the constant C depends on
a specific point of the line.
r
The vectors n = (−u y , u x ) = ( A, B ) or
r
n = (u y ,−u x ) = ( A, B ) are perpendicular to the vector
r
u and so they are perpendicular to the line L .
These are called normal vectors to the line L .
Let P( x, y ) be a generic point on the line L . So:
r
r
r
P0 P || u ⇒ P0 P ⊥ n ⇒ P0 P ⋅ n = 0
r r r
(r − r0 ) ⋅ n = 0
The normal equation of a line is given by:
r r r
(r − r0 ) ⋅ n = 0
8.2 Cartesian Equation of a Line
©2010 Iulia & Teodoru Gugoiu - Page 1 of 2
Calculus and Vectors – How to get an A+
Ex 3. Convert the vector equation of the line
r
L : r = (1,2) + t (3,−2), t ∈ R to the Cartesian
equation.
r
r
u = (3,−2) ⇒ n = (2,3) = ( A, B )
Ex 4. Convert the Cartesian equation to the parametric
equations and then to the vector equation.
L : −x + 2 y + 3 = 0
⎧ x = 3 + 2t
, t∈R
⎨
⎩y = t
r
r = (3,0) + t (2,1), t ∈ R
2x + 3y + C = 0
P0 (1,2) ∈ L ⇒ 2(1) + 3(2) + C = 0 ⇒ C = −8
∴ L : 2x + 3y − 8 = 0
D Slope y-intercept Equation
Let solve the symmetric equation of a line:
x − x0 y − y 0
=
= t, t ∈ R
ux
uy
for y :
y = −2 x + n
(−2,3) ∈ L ⇒ 3 = (−2)(−2) + n ⇒ n = −1
∴ y = −2 x − 1
x − x0
y − y0 = u y
ux
y=
uy
x + y0 −
Ex 5. Convert the vector equation to the slope y-intercept
r
equation: L : r = (−2,3) + t (1,−2)
uy − 2
r
u = (1,−2) ⇒ m =
=
= −2
1
ux
uy
x0
ux
ux
The slope y-intercept equation of a line in R2 is
given by:
y = mx + n
uy
m=
ux
Ex 6. Convert the slope y-intercept equation to the vector
equation.
2
L: y = − x+4.
3
m=
where m is the slope and n is the y-intercept
which depends on a specific point of the line.
E Angle between two Lines
The angle between two lines is determined by the
angle between the direction vectors:
r r
u ⋅u
cosθ = r 1 r2
|| u1 || || u 2 ||
Note. There are two pairs of equal angles
between two lines (see the figure below).
r
− 2 uy
=
⇒ u = (3,−2)
3
ux
x = 0 ⇒ y = 4 ⇒ (0,4) ∈ L
r
L : r = (0,4) + t (3,−2), t ∈ R
Ex 7. Find the acute angle between each pair of lines.
r
x − 2 y +1
=
a) L1 : r = (0,1) + t (1,−3), t ∈ R; L2 :
3
−2
r
r
u1 = (1,−3), u 2 = (3,−2)
r r
u ⋅u
3+ 6
9
cosθ = r 1 r2 =
=
|| u1 || || u 2 ||
1+ 9 9 + 4
10 13
9
∴θ = cos −1
≅ 37.87°
10 13
1
b) L1 : 2 x − 3 y + 6 = 0; L2 : y = − x + 2
3
r
r
n1 = (2,−3) ⇒ u1 = (3,2)
Note:
u2 y
r
1
=
⇒ u 2 = (−3,1)
− 3 u2x
r r
u ⋅u
−7
−7
=
⇒ θ = cos −1
≅ 127.87°
cos θ = r 1 r2
|| u1 || || u 2 ||
13 10
13 10
∴ θ1 = 180° − θ = 180° − 127.87° ≅ 52.13°
m2 =
θ1 + θ 2 = 180°
Reading: Nelson Textbook, Pages 435-442
Homework: Nelson Textbook: Page 443 #1, 3, 4, 5, 6, 7, 8, 10ab, 11, 14
8.2 Cartesian Equation of a Line
©2010 Iulia & Teodoru Gugoiu - Page 2 of 2