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Illustrative
Mathematics
F-LE Do two points always
determine a linear function II?
Alignments to Content Standards: F-LE.A.2
Task
Suppose R1
= (c, d) and R2 = (e, f ) are two different points.
a. Is there always a linear function f (x)
and R2 ? Explain.
= mx + b whose graph contains the points R1
b. Find a linear equation of the form ax + by
solutions to the equation.
= k so that (1, 6) and (4, −3) are both
c. Find a linear equation of the form ax + by
solutions to the equation.
= k so that (2, 8) and (2, −3) are both
d. Show that there is always a linear equation of the form ax + by
and (e, f ) are both solutions to the equation.
= k so that (c, d)
IM Commentary
This task is designed as a follow-up to the task F-LE Do Two Points Always Determine a
Linear Function? Linear equations and linear functions are closely related, and there
advantages and disadvantages to viewing a given problem through each of these
points of view. For example, all lines can be described by linear equations, while only
non-vertical lines can be described as the graph of a linear function. On the other hand,
the values of m and b in a linear function of the form f (x) = mx + b are completely
determined by any two points on its graph, whereas the values of a, b, and k in the
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linear equation ax + by = k are not: a, b and k can all be replaced with, for example,
2a, 2b, and 2k without changing the solutions to the equation.
This relationship between linear equations and linear functions studied here is closely
related to the relationship between ratios and rational numbers and this arises in one
of the solutions to the problem. Indeed lines through the origin represent the set of
pairs (x, y) sharing the ratio (x : y). These lines can also be described by the slope y/x
provided the line is not vertical. So the slope describes a non-vertical line through (0, 0)
with a single number, as long as the line is not vertical, whereas the ratio (x : y) can be
used to describes any line through the origin, at the cost of uniqueness of
representation; (x : y) describes the same line as (rx : ry) for any non-zero real
number r.
A quick walk-through of the solutions: The first solution given is complete, using a
geometric argument to argue the existence of a line through two properties. We note
that an algebraic construction of the coefficients for this line is more difficult, requiring
patience and good algebra skills from the students. Solution (b) is included to give such
an algebraic version of this deduction, giving an explicit construction of a linear
equation for such a line. Solutions (c) and (d) are included to detail the procedure in
the case cf − de = 0 via different approaches: Solution (c) does so through the use of
ratios, whose relevance was introduced in the previous paragraph, and solution (d)
takes a more algebraic approach.
This task is not intended for assessment purposes: rather it is intended to show the
depth of the standard F-LE.2 and its relationship to other important concepts of the
middle school and high school curriculum, including ratio, algebra, and geometry.
Solutions
Edit this solution
Solution: 1 Algebra and geometric solution to part (d)
a. As was seen in the problem "Do two points always determine a linear function?'' if
c = e but d ≠ f then there cannot be a linear function f (x) = mx + b whose graph
contains both (c, d) and (e, f ). In part (b) of that problem the case (c, d) = (0, 5) and
(e, f ) = (0, 7) was considered. No such function f (x) can exist because the y-intercept,
b, would have to be simultaneously equal to 5 and to 7. Alternatively, plugging in (0, 5)
to the equation f (x) = mx + b gives b = 5 while plugging in (0, 7) to the equation
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f (x) = mx + b gives b = 7 and these two equations are inconsistent.
b. Plugging the points
equations
(1, 6) and (4, −3) into the equation ax + by = k give the two
a + 6b = k
4a − 3b = k.
Multiplying the second equation by 2 and then adding this to the first equation gives
9a = 3k.
We do not have enough information to solve for both a and k but we may choose a
convenient value of k and then use this information to solve for a and b. If we choose
k = 3 then we find that a = 1. Plugging these values into a + 6b = k gives us that
b = 1 . So the graph of the equation
3
x+
1
y=3
3
contains the two points (1, 6) and (4, −3). If we were to choose k
instead of k
= 3, we can solve for a and b and find that a =
different equation of
= 5 for example,
b = 59 giving us a
5
and
3
5
5
x + y = 5.
3
9
There are two important facts to notice about these two equations. First, the second
equation is found by multiplying the first equation by 5 . Secondly, if the two equations
3
are solved for y in terms of x, they both give the same equation, namely y = −3x + 9.
This is one of the reasons why the slope/intercept form for the equation of a line is
valuable because there can only be one such equation whose graph describes a given
line.
c. For the points (2, 8) and (2, −3) note that the x-coordinate of both points is 2 and so
these points like on the vertical line x = 2. Like in part (b) above, this equation is not
the only one which will work: 2x = 4 also contains (2, 8) and (2, −3).
d. For part (d) we know from Euclidean geometry that given two distinct points R1 and
R2 in R2 there is a unique line L that contains R1 and R2 . A line L in R2 is either
vertical, in which case it can be described by an equation of the form x = c where c is a
real number, or it is not vertical and then it can be described as the solutions to an
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equation of the form y = mx + b where m is the slope and b is the y-intercept. This can
be put it into the required form by re-arranging as (−m)x + (1)y = b.
It is important to stress the logic here: this geometric approach only shows that there is
an equation of the form ax + by = k whose solutions contain R1 and R2 . Finding
suitable a, b, and k as in the above algebraic solution requires substantially more work
and this is presented in the second solution.
Edit this solution
Solution: 2. One algebraic solution to part (d)
In order for (c, d) to be a solution to ax + by
= k we must have
ac + bd = k.
Similarly for (e, f ) to be a solution to ax + by
= k we need
ae + bf = k.
We would like to manipulate these equations to solve for a and b. We have two
equations in two unknowns so we would like to eliminate one variable and then solve
for the other. We will try to eliminate a first. We can do this by multiplying the first
equation by e, multiplying the second equation by c and then subtracting these two
equations. This gives us, subtracting the second scaled equation from the first,
b(de − cf ) = k(e − c).
We can now solve for b, provided that de − cf is not zero. We will assume that de − cf
is non-zero and then deal later with the case when it is zero. We find
b=
k(e − c)
.
de − cf
We can solve for a using the same method and we find
a=
We may now choose k
k(f − d)
.
cf − de
= 1 and we have the equation
f −d
e−c
x+
y=1
( cf − de )
( de − cf )
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whose graph contains the two points (c, d) and (e, f )
Since (c, d) and (e, f ) are distinct points, not all of c, d, e, f are zero. For simplicity,
assume c ≠ 0. (There are analogous cases if c = 0 but one of the others is not.) Since
c ≠ 0, the equation −dx + cy = 0 defines a line which contains the point (c, d) as we
verify by substituting x = c and y = d. If we substitute x = e and y = f we get the
equation −de + cf = 0 which is true by our assumption. Therefore the line defined by
the equation −dx + cy = 0 contains (c, d) and (e, f ) when cf − de = 0.
While this is a reasonable place to conclude an algebraic response to part (d), we
include in the following two solutions different approaches of more fully addressing the
case that cf − de = 0.
Edit this solution
Solution: Extended solution to the cf
− de = 0 case in (d) using ratios
We give here an approach to the case where cf − de = 0 using the idea of ratio and
proportion. The relation cf − de = 0 means that the two ratios (c : d) and (e : f ) are
equivalent, which implies that the points (c, d) and (e, f ) are both contained on a line
through the origin. An equation defining this line is given by −dx + cy = 0.
Alternatively, using the point (e, f ) we find the equation −f x + ey = 0. Both equations
define the same line because (c : d) and (e : f ) are equivalent.
A more concise way to explain the above is as follows: The line −dx + cy = 0 always
contains the point (c, d) as we verify by substituting x = c and y = d. If we substitute
x = e and y = f we get the equation −de + cf = 0, which is precisely the case we're
addresing. Therefore the line defined by the equation −dx + cy = 0 contains (c, d) and
(e, f ) when cf − de = 0.
Edit this solution
Solution: Extended solution to the cf
cases
− de = 0 case in (d) by breaking into
In this solution, we check algebraically what happens when cf − de = 0. We divide this
into two cases: first when none of the numbers c, d, e, f are zero, we will check that the
equivalent equations −dx + cy = 0 and −f x + ey = 0 define the line through (c, d)
and (e, f ). We then show that even when only one of c, d or one of e, f are non-zero
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these equations still define the desired line, −dx + cy
−f x + ey = 0 in the second.
= 0 in the first case and
If none of c, d, e, f are zero, the relation cf − de = 0 leads to a number of different
equations between rational expressions which give us the equation of our line
containing (c, d) and (e, f ). For example, we have dc = e and so the line through the
origin with slope dc
e
contains both points
f
f
=
(c, d) and (e, f ). This line can be written in
the form −dx + cy = 0 or −f x + ey = 0 and so we have found the desired equation
when none of c, d, e, f are zero. It turns out, as we will show next, that these equations
work in all cases but we need to be careful to choose one where the pair of coefficients
(−d and c for the first, −f and e for the second) are not both equal to zero.
We will now see what happens when one or more of the numbers c, d, e, f is zero.
Remember that we are assuming cf − de = 0. So if one of the quantities, say c, is zero,
then this forces either d or e, or both, to be zero.
a. Case 1: c = 0
i. subcase 1a: e
= 0. The line x = 0 is required equation.
ii. subcase 1b: d
= 0. We are looking at a line through the origin and the point
f
(e, f ); the equation is y = e x, which can be written −f x + ey = 0.
b. Case 2: f
=0
i. subcase 2a: e
= 0. We are looking at line through the origin and the point (c, d);
the equation is y = dc x, which can be written −dx + cy = 0.
ii. subcase 2b: d
= 0. The line y = 0 is the required equation.
We can proceed through four similar cases (with some duplication of effort), addressing
the situations when d or e is zero.
F-LE Do two points always determine a linear function II?
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