Projects - Cengage Learning

C-84 Appendix C
y
(0, 5)
P(x, y)
Figure B
P 0
(_2, 1)
tA
”_ 5 , 0
2
P
”
0
P
A= 1, 2
x
Equation (3) is called the symmetric equation for this line. If either a or b is
zero then the line has no symmetric equation.
Let’s apply equations (1), (2), and (3) to find vector, parametric, and
symmetric equations for the line in the x-y plane passing through the point
P 0 (2, 1) with direction A 81, 29. See Figure B. (Note: This is a continuation
of Example 1 from the previous project on lines, circles, and ray tracing.)
A vector equation for this line is
Then
P P 0 tA for q t q
8x, y9 82, 19 t 81, 29 for q t q
8x, y9 82 t, 1 2t9 for q t q
And we obtain parametric equations for the line
x 2 t
b
y 1 2t
Let’s use the parametric equations to find the x and y-intercepts of the line.
For the x-intercept, y 0 implies 1 2t 0 or t 12. Then the x-intercept
is x 2 12 52. For the y-intercept, x 0 implies 2 t 0 or
t 2. Then the y-intercept is y 1 2(2) 5. See Figure B. Finally, eliminating
the parameter t in the parametric equations gives us the symmetric
equation for this line,
x 2
1
From the symmetric equation we have
the slope-intercept equation for this line.
for q t q
y 1
2
x 2 y 1
2
2(x 2) y 1
y 2x 5