Do two points always determine a linear function II?
Task
Suppose $R_1 = (c,d)$ and $R_2 = (e,f)$ are two different points.
 Is there always a linear function $f(x) = mx + b$ whose graph contains the points $R_1$ and $R_2$? Explain.
 Find a linear equation of the form $ax + by = k$ so that $(1,6)$ and $(4,3)$ are both solutions to the equation.
 Find a linear equation of the form $ax + by = k$ so that $(2,8)$ and $(2 ,3)$ are both solutions to the equation.
 Show that there is always a linear equation of the form $ax + by = k$ so that $(c,d)$ and $(e,f)$ are both solutions to the equation.
IM Commentary
This task is designed as a followup to the task FLE 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 nonvertical 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 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 nonvertical 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 nonzero real number $r$.
A quick walkthrough 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 $cfde=0$ 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 FLE.2 and its relationship to other important concepts of the middle school and high school curriculum, including ratio, algebra, and geometry.
Solutions
Solution: 1 Algebra and geometric solution to part (d)
 As was seen in the problem "Do two points always determine a linear function?'' if $c = e$ but $d \neq f$ 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 $f(x) = mx + b$ gives $b = 7$ and these two equations are inconsistent.

Plugging the points $(1,6)$ and $(4,3)$ into the equation $ax + by = k$ give the two equations \begin{eqnarray*} a + 6b &=& k \\ 4a  3b & = & k. \end{eqnarray*} 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 $b = \frac{1}{3}$. So the graph of the equation $$ x + \frac{1}{3}y = 3 $$ contains the two points $(1,6)$ and $(4,3)$. If we were to choose $k = 5$ for example, instead of $k = 3$, we can solve for $a$ and $b$ and find that $a = \frac{5}{3}$ and $b = \frac{5}{9}$ giving us a different equation of $$ \frac{5}{3}x + \frac{5}{9}y = 5. $$ There are two important facts to notice about these two equations. First, the second equation is found by multiplying the first equation by $\frac{5}{3}$. Secondly, if the two equations 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.
 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 $(2,3)$.

For part (d) we know from Euclidean geometry that given two distinct points $R_1$ and $R_2$ in ${\bf R}^2$ there is a unique line $L$ that contains $R_1$ and $R_2$. A line $L$ in ${\bf R}^2$ 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 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 rearranging 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 $R_1$ and $R_2$. Finding suitable $a$, $b$, and $k$ as in the above algebraic solution requires substantially more work and this is presented in the second 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, $$ b(decf) = k(e  c). $$ We can now solve for $b$, provided that $de  cf$ is not zero. We will assume that $de  cf$ is nonzero and then deal later with the case when it is zero. We find $$ b = \frac{k(e  c)}{de  cf}. $$ We can solve for $a$ using the same method and we find $$ a = \frac{k(fd)}{cfde}. $$ We may now choose $k = 1$ and we have the equation $$ \left(\frac{fd}{cfde}\right)x + \left(\frac{e  c}{de  cf}\right)y = 1 $$ 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 $c\neq 0$. (There are analogous cases if $c=0$ but one of the others is not.) Since $c\neq 0$, 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 $cfde=0$.
Solution: Extended solution to the $cfde=0$ case in (d) using ratios
We give here an approach to the case where $cfde=0$ 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 $fx + ey = 0$. 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 $cf  de = 0$.
Solution: Extended solution to the $cfde=0$ case in (d) by breaking into cases
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 $fx +ey =0$ 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 nonzero these equations still define the desired line, $dx + cy = 0$ in the first case and $fx + ey = 0$ in the second.
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 $\frac{d}{c} = \frac{e}{f}$ and so the line through the origin with slope $\frac{d}{c} = \frac{e}{f}$ contains both points $(c,d)$ and $(e,f)$. This line can be written in the form $dx + cy =0$ or $fx +ey =0$ 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.
 Case 1: $c =0$
 subcase 1a: $e=0$. The line $x=0$ is required equation.
 subcase 1b: $d =0$. We are looking at a line through the origin and the point $(e,f)$; the equation is $y = \frac{f}{e}x$, which can be written $fx + ey =0$.
 Case 2: $f =0$
 subcase 2a: $e=0$. We are looking at line through the origin and the point $(c,d)$; the equation is $y = \frac{d}{c}x$, which can be written $ dx + cy =0$.
 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.
Do two points always determine a linear function II?
Suppose $R_1 = (c,d)$ and $R_2 = (e,f)$ are two different points.
 Is there always a linear function $f(x) = mx + b$ whose graph contains the points $R_1$ and $R_2$? Explain.
 Find a linear equation of the form $ax + by = k$ so that $(1,6)$ and $(4,3)$ are both solutions to the equation.
 Find a linear equation of the form $ax + by = k$ so that $(2,8)$ and $(2 ,3)$ are both solutions to the equation.
 Show that there is always a linear equation of the form $ax + by = k$ so that $(c,d)$ and $(e,f)$ are both solutions to the equation.