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Straight Lines
An equation of the form
Ax + By = C
is called a “linear equation” because the set of all points (x, y)
that satisfy the equation is a straight line in the plane.
This form of the equation of a straight line is called the “general
equation” of a straight line.
If B = 0, the line has the form x = a, which is a vertical line.
In this case, it is not the graph of a function. Otherwise, there
are many other useful forms of the straight line equation, all of
which define y as a function of x.
(x , y )
2 2
y − y =∆y
y-intercept b
(x , y )
1 1
x-intercept a
The slope of the line is m =
θ
x − x =∆x
2
y2 − y1
x2 − x1
2
1
∆y
=
∆x
= tan(θ )
1
1. If B is not 0, we can solve for y and get
y =− A x + C or y = mx + b
B
B
This is called the “slope intercept” form of the equation,
because it involves the slope m and the y-intercept b.
2. Now suppose that we know the slope m and one point
(x , y ) on the line.
1 1
Then if (x, y) is an arbitrary point on the line, we also
see that the slope can also be written as:
y− y
m= 1
x− x
1
So that
( y − y ) = m(x − x )
1
1
This is called the “point-slope” form of the equation of a line
3. If we know two points on the line, we can find the slope
using the formula:
y2 − y1
∆
y
m= =
∆x x2 − x1
And then substituting this into the point-slope formula.
4. Finally, if the constant term is not 0, we can divide by it and
write the general equation in the “double-intercept” form:
x y
+ =1
a b
Problem 1. Find the slope of the line segment from the point (1, 2)
to the point (− 1, 4).
Solution: Let (−1, 4) be the second point (arbitrarily). Then ∆y/∆x
= 2/ − 2 = −1.
Problem 2. Determine if the points (1,1), (2, 3), and (4, 7) lie on
the same line.
Solution: This is true if and only if the slope of the line segment
from the first to the second point is the same as the slope of the
segment from the first to the third point.
(x , y )
2 2
(x , y )
3 3
(x , y )
1 1
The slope of the line segment between the points (1,1) and (2, 3)
is (3 − 1)/(2 − 1) = 2, and the slope of the line segment the points
(1,1) and (4, 7) is 6/3 = 2. These are equal, so the three points lie
on the same line.
Problem 3. Find the equation of the line whose slope is 3 and
which passes through the point (5, 4).
Solution: Using the point slope form of the equation, we have
y − 4 = 3(x − 5), or y = 3x − 11.
Problem 4. Find the equation of the line passing through the
two points (1, 5) and (4, 2).
Solution: The slope is the change in y over the change in x, or
(2 − 5)/(4 − 1) = −1. If we choose the first point for the pointslope form, we have y − 5 = −1(x − 1), or y = −x + 6.
Theorem: Two lines are parallel if and only if they have the
same slope
Parallel lines have the same
slope.
Theorem. Two lines are perpendicular if and only if the product
of their slopes is -1.
Perpendiculars have negative
reciprocal slopes
a c
α
s
b
t
β
The slopes of the two lines are c/s and –c/t respectively. Now
we see:
a2 +b2 = (s + t)2 = s2 + 2st + t 2
a2 = c2 + s2
Thus a2 +b2 = s2 + 2c2 + t 2 and so c2 = st
and the product of the two slopes is –1.
b2 = c2 +t 2
Problem 5. Find the equation of the line that is parallel to the line
with equation y = 4x – 2 and that has y-intercept 7.
Solution. The required line has slope 4 and y – intercept 7, so its
equation is y = 4x + 7
Problem 6. Find the equation of the line that passes through the
point (3, -4) and is perpendicular to the line with equation x – 4y = 7.
Solution. The equation x – 4y = 7 can be rewritten as y = x/4 – 7/4,
so its slope is ¼. The slope of the required line is therefore –4. The
point slope form is (y+ 4) = – 4(x – 3) , or y = – 4x + 8 .
Problem 7. In each case classify the lines as parallel, perpendicular,
or neither.
(a) y = – 5x + 1
y = 3 – 5x
(b) (y – 1) = 2(x – 3)
(y – 4) = (– 1/2)(x + 7)
(c) 4x + 5y + 7 = 0
5x – 4y + 9 = 0
(d) Ax + By + C = 0
Ax + By + D = 0
(e) y = (1/2)x
x = (1/2)y
Solution. (a)Slopes are – 5 and –5, so lines are parallel
(b) Slopes are 2 and –1/2, so lines are perpendicular
(c) Slopes are – 4/5 and 5/45, so lines are perpendicular
(d) Slopes are – A/B and – A/B , so lines are parallel
(e) Slopes are 1/2 and 2, so lines are neither.
If y = mx + b is the formula for a function, then we say that y
varies linearly with x. In this case y has a constant rate of change
with respect to x, namely m.
This means that if x changes by any amount ∆x, then y always
changes by the amount ∆y =m∆x. This is true since the equation
in point-slope form is
( y − y ) = m(x − x )
1
1
and so at any other point we have
( y − y ) = m(x − x ) or ∆y=m∆x
2
1
2
1
Many practical models of real phenomena are linear.
Examples:
1. An object moves on a straight line with constant velocity v. If
the initial position is s0, and the position at any time t is s(t),
then we have s(t) = vt + s0. Thus position is a linear function of
time.
2. The length y of a spring is a linear function of the weight x
attached to it.
y
y = kx + y0
wt = x
3. A metal rod of length 60 cm. has its ends held at constant
temperatures 5 degrees C and 30 degrees C. Find an expression
for the temperature at any point of the rod.
Locate the left hand end of the rod at 0 on a line and the right
hand end at 60. At equilibrium, the temperature at any point x
on the rod is a linear function of x.
T = mx +b
Since the T = 5 at x = 0, the T-intercept is 5. The slope is
computed as:
m = ∆T = 55
∆x
60
Thus T = (55/60)x + 5.
0
60
The figure shows position versus time for a particle moving along
the x-axis.
(a) What is the velocity of the particle?
(b) What is the x-coordinate of the particle at t = 0?
(c) What is the x-coordinate of the particle at t = 5?
(d) At what time does the particle have x-coordinate 2?