Question

linear function

Answer 1

Given two distinct points (a,b) and (c,d), find the linear function f whose graph goes through (a,b) and (c,d) amounts to saying that f(a) = b and f(c) = d
If f is to be of the form
\[f(x) = \alpha a+\beta\]
then we have
\[\alpha a+\beta =b\]
\[\alpha c+\beta =d; \]
therefore
\[\alpha =\frac{ d-b }{ c-a }\]
and \[\beta=b-\left[ \left( \frac{ d-b }{ c-a } \right) \right]a\]
so,
\[f(x) = \frac{ d-b }{ c-a }x+b-\frac{ d-b }{ c-a }a=\frac{ d-b }{ c-a }(x-a)+b

Answer 2

\[f(x) = \frac{ d-b }{ c-a }x+b-\frac{ d-b }{ c-a }a=\frac{ d-b }{ c-a }(x-a)+b\]

Answer 3

that's like the most complicated point-slope form I've seen…
I just need the alpha and beta portion to be elucidated.

Answer 4

\[a \neq 0\]

Answer 5

That's a neat derivation. And why do we need \(a\neq0\) in this?

Answer 6

sorry it's \[a \neq c\]

Answer 7

which accounts only for the straight-line not parallel to the vertical-axis

Answer 8

So you just wanted to clarify how they solved for \(\alpha,\beta\)?

Answer 9

yes and like what are those supposed to represent?

Answer 10

In this case, \(\alpha\) and \(\beta\) are just two different real numbers.

As for solving for them. You start with the two equations\[\alpha a+\beta =b\implies\beta=\alpha a-b\]\[\alpha c+\beta =d\implies\beta=\alpha c-d.\]Thus,\[\alpha a-b=\alpha c-d\implies\alpha c-\alpha a=d-b\]Factor, the \(\alpha\) out, and we get\[\alpha=\frac{d-b}{c-a}.\]