Projects - Cengage Learning

Appendix C C-79
y
y=2x+5
Îy=2
A
(1, 2)
P 0
A
(_2, 1)
Îx=1
Îy=2
Îx=1
x
Figure B
(b) Let P (x, y) be any point on the line and let P 0 (2, 1). So the position
vectors from the origin to these points are P 8x, y9 and P 0 82, 19.
Given the slope 2, we can find the line’s direction by drawing the line of
slope 2 passing through the origin and picking a convenient point (different
from the origin) on this line. See Figure B. Using the point A (1, 2),
we see that the vector A 81, 29 is parallel to the line. So a vector equation
for the line is
P P 0 tA for q t q
which in component form becomes
8x, y9 82, 19 t81, 29 for q t q
(c) For t 1, 8x, y9 82, 19 (1) 81, 29 83, 19. So the vector
83, 19 goes from the origin to the point P on the line, with P (3, 1).
Similarly, for t 0 the point on the line is P (2, 1) and for t 2 the
point is P (0, 5).
Exercise 1 Consider the line in the x-y plane passing through the point
(1, 2) with slope 3.
(a) Sketch the line and find its slope-intercept equation.
(b) Find a vector equation for the line. Label three vectors on the line in
part (a).
(c) Find the points on this line corresponding to t 2, 1, 0, 1, and 2.
Exercise 2 Consider the line in the x-y plane passing through the points
A (3, 1) and B (2, 4).
(a) Sketch the line and then find a vector equation for it.