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Method 2:
We start by writing the given line in parametric form, which is
1x, y2 13, 62 s11, 42 or 1x, y2 13 s, 6 4s2. This gives the
parametric equations x 3 s and y 6 4s. To find the required equation,
we solve for s in each component. Thus, s x 3 and s y 6 Since these
1
4 .
equations for s are equal,
x 3
1 y 6
4
y 4x 18
Therefore, the required equation is y 4x 18, which is the same answer we
obtained using Method 1. The graph of this line is shown below.
y
4
–2 0
–4
m = (– 1, – 4)
–8
–12
–16
–20
x
2 4 6
(3, –6)
y = 4x – 18
41x 32
1
y 6
y 6 41x 32
In the example that follows, we examine the situation in which the direction
vector of the line is of the form m ! 10, b2.
EXAMPLE 3
Reasoning about equations of vertical lines
Determine the Cartesian form of the line with the equation
r ! 11, 42 s10, 22, sR.
Solution
The given line passes through the point 11, 42, with direction vector 10, 22, as
shown in the diagram below.
6
y
4 A(1, 4)
2
m = (0, 2) B(1, 0) x
–4 –2 0 2 4
–2
–4
NEL
CHAPTER 8 437