Page 27 - Calculus Demystified

CHAPTER 1
Basics
14
You Try It: Write the equation of the line that passes through the point (−3, 2)
and has slope 4.
EXAMPLE 1.11
Write the equation of the line passing through the points (−4, 5) and (6, 2).
SOLUTION
Let (x, y) be a variable point on the line. Using the points (x, y) and (−4, 5),
we may calculate the slope to be
y − 5
m = .
x − (−4)
On the other hand, we may use the points (−4, 5) and (6, 2) to calculate the
slope:
2 − 5 −3
m = = .
6 − (−4) 10
Equating the two expressions for slope, we find that
y − 5 −3
= .
x + 4 10
Simplifying this identity, we find that the equation of our line is
−3
y − 5 = · (x + 4).
10
You Try It: Find the equation of the line that passes through the points (2, −5)
and (−6, 1).

In general, the line that passes through points (x 0 ,y 0 ) and (x 1 ,y 1 ) has equation
y − y 0 y 1 − y 0
= .
x − x 0 x 1 − x 0
This is called the two-point form of the equation of a line.
EXAMPLE 1.12
Find the line perpendicular to y = 3x − 6 that passes through the point
(5, 4).
SOLUTION
We know from the Math Note immediately after Example 1.10 that the given
line has slope 3. Thus the line we seek (the perpendicular line) has slope −1/3.
Using the point-slope form of a line, we may immediately write the equation
of the line with slope −1/3 and passing through (5, 4) as
−1
y − 4 = · (x − 5).
3