3.4 The Point-Slope Form of a Line

298 Chapter 3 Linear Functions
P (−2,2)
Version: Fall 2007
Q(−3,−1)
y
R(2,1)
(a) The line through
Q(−3, −1) and R(2, 1).
x
Figure 4.
P (−2,2)
Q(−3,−1)
y
∆x=5
T (3,4)
R(2,1)
∆y=2
(b) The line through P (−2, 2)
that is parallel to the
line through Q and R.
We now draw a line through the point P (−2, 2) that is parallel to the line through
the points Q and R. Parallel lines must have the same slope, so we start at the point
P (−2, 2), “run” 5 units to the right, then “rise” 2 units up to the point T (3, 4), as
shown in Figure 4(b).
We seek the equation of the line through the points P and T . We’ll use the pointslope
form of the line
y − y0 = m(x − x0). (13)
We’ll use the point P (−2, 2) as the given point (x0, y0). That is, (x0, y0) = (−2, 2).
The line through P has slope 2/5. Substitute m = 2/5, x0 = −2, and y0 = 2 in
equation (13) to obtain
y − 2 = 2
(x − (−2)). (14)
5
Let’s place the equation (14) in standard form. Distribute the slope, then clear
fractions by multiplying both sides of the resulting equation by 5.
y − 2 = 2 4
x +
5� 5 �
2 4
5(y − 2) = 5 x +
5 5
5y − 10 = 2x + 4
Finally, subtract 5y from both sides of the last equation, then subtract 4 from both
sides of the equation, obtaining
or equivalently,
−14 = 2x − 5y,
x