Math 141 Lecture Notes for Section 1.4
Section 1.4 -
1
Intersection of Straight Lines
Suppose we have two lines L1 and L2 with equations
y = m 1 x + b1
and
y = m 2 x + b2
respectively. Furthermore then suppose that the two lines intersect at a point (x0 , y0 ). The point lies
on L1 so we have that
y0 = m1 x0 + b1
and it lies on L2 so
y0 = m2 x0 + b2
What these equations tell us is that any point where L1 and L2 has x value such that
m1 x + b1 = m2 x + b2
and we find the y coordinate of the point by evaluating either of the linear equations at x. It is important
to note here that any pair of linear equations has either no solutions, one solution or infinitely many
solutions. The first case happens when two lines don’t intersect, which can only happen if both the
lines are parallel and non-equal. The third case can happen only if both lines are exactly the same line.
We can test for these two cases by simply checking if m1 = m2 . If they are equal, then they are the
same line if b1 = b2 and have infinitely many solutions. If the two intercepts aren’t equal, then the lines
are parallel and non-intersection so no solutions exist.
Example 1.4.1 (Intersection of Two Lines):
Consider lines L1 and L2 represented by the equations
4x + 2y − 6 = 0
and
y = 3x + 5
respectively and find their intersections.
First we need to change the equation
4x + 2y − 6 = 0
into slope-intercept form. We first divide both sides of the equation by 2 and we have
2x + y − 3 = 0.
Then we subtract 2x and add 3 and get
y = −2x + 3
. Since we now know that L1 has slope −2 and L2 has slope 3, there must be exactly one solution.
Setting both equations for y equal to each other yields the following steps:
3x + 5 = −2x + 3
3x + 5 + 2x = −2x + 3 + 2x
5x + 5 = 3
5x + 5 − 5 = 3 − 5
5x = −2
x = − 52 .
We now know the x-coordinate of the point of intersection is − 25 and we can compute that the
y-coordinate must be 19
5 .
Math 141 Lecture Notes for Section 1.4
2
Example 1.4.2 (Break-Even Analysis):
Suppose you decide to open a sub shop. You have a fixed operating cost of $55, 000 per month, a unit
cost of $1.50 per sandwich and a unit price of $6.00 per sandwich. How many subs do you need to sell
in order to break even?
From Section 1.3 we know the cost equation C(x) for selling x sandwiches is
C(x) = mx + b
where m is the unit cost 1.5 and b is the fixed cost, 55, 000. This gives us the equation
C(x) =
3
x + 55, 000
2
and the revenue equation is
R(x) = sx
where s is the unit price, 6, so
R(x) = 6x
We now set the equations equal to one another and have
3
2x
3
2x
+ 55000 = 6x
+ 55000 − 23 x = 6x − 23 x
55000 = 92 x
110000
=x
9
This divides out to be approximately 12, 223 sandwiches to break even.
Example 1.4.3 (Market Equilibrium):
A textbook company is trying to sell math textbooks. It is determined that if the unit price is $125
then the quantity demanded is 100, 000 and if the unit price is $75 the quantity demanded is 300, 000.
Similarly, if there are 600, 000 textbooks supplied the unit price is $90 and if there are 200, 000 textbooks
supplied the unit price is $40. Find the market equilibrium for this math textbook.
First we need to find the linear equations for both suppy and demand. When calculating suppy and
demand it is important to note that we have quantity as the independent variable and price as the
dependent variable.
First we compute the slope of the demand function using the given points of (100000, 125) and
(300000, 75), this gives us slope
m1 =
125 − 75
50
1
=−
=−
.
100000 − 300000
200000
4000
Using the point-slope form of a line
1
p − 125 = − 4000
(x − 100000)
1
p − 125 = − 4000 x + 25
1
p = − 4000
x + 150
Math 141 Lecture Notes for Section 1.4
3
Now for the supply function, we have been given points of (600000, 90) and (200000, 40), then we have
slope
90 − 40
50
1
m2 =
=
=
600000 − 200000
4000000
8000
and use the point-slope form of the line to obtain
p − 40 =
p − 40 =
p =
1
8000 (x − 200000)
1
8000 x − 25
1
8000 x + 15
Setting both equaitons equal to each other gives us
1
8000 x
1
+ 15 = − 4000
x + 150
1
1
x
=
−
8000
4000 x + 135
3
x
=
135
8000
x = 360, 000.
Thus the point of market equilibrium is at 360, 000 textbooks with a price of $60.
Suggested Homework Problems: 3,9,13,21,25,27.