Change of interval intuition? Suppose $t \in [c,d]$. Then we can define a linear function $\lambda(t)$ s.t. $\lambda(t) \in [a,b]$. That is, we can convert one interval to another interval. This function can be defined by $\lambda(t) = \dfrac{b-a}{d-c}t + \dfrac{a\cdot d-b\cdot c}{d-c}$.
What is the intuition for this formula? I get the first part: If for example the interval $[c,d]$ is double the size of $[a,b]$, then $\cfrac{b-a}{d-c} = 0.5$ to keep $\lambda(t)$ within the bounds. But why does the second part work/how did we end up with that fraction?
 A: hint
We look for a linear transformation of the form $$\lambda(t)=\alpha t+\beta$$
which takes $ c $ to $ a$ and $ d $ to $ b$ :
$$\lambda(c)=\alpha.c+\beta = a$$
$$\lambda(d)=\alpha.d+\beta = b$$
the difference gives
$$\alpha(d-c)=(b-a)$$
or
$$\alpha=\frac{b-a}{d-c}$$
replacing $ \alpha$ , we get $\beta$.
or we can eliminate $\alpha$ as
$$d\lambda(c)-c\lambda(d)=ad-bc$$
$$=\beta(d-c)$$
A: It’s simply the equation of the straight line through the points $\langle c,a\rangle$ and $\langle d,b\rangle$: the segment of that line between those two endpoints is clearly the graph of a bijection from $[c,d]$ (on the $x$-axis) to $[a,b]$ (on the $y$-axis).
The slope of the line is $\frac{b-a}{d-c}$, and it passes through $\langle c,a\rangle$, so we have
$$y-a=\frac{b-a}{d-c}(x-c)\,,$$
or
$$\begin{align*}
y&=\frac{b-a}{d-c}x-\frac{c(b-a)}{d-c}+a\\
&=\frac{b-a}{d-c}x+\frac{a(d-c)-c(b-a)}{d-c}\\
&=\frac{b-a}{d-c}x+\frac{ad-bc}{d-c}\,.
\end{align*}$$
A: From $t$ in $[c,d]$, we can deduce a reduced variable in the unit range $[0,1]$:
$$\mu(t):=\frac{t-c}{d-c}.$$
And from a variable in $[0,1]$, we deduce a variable in $[a,b]$:
$$\lambda(t)=(b-a)\mu(t)+a=(b-a)\frac{t-c}{d-c}+a.$$
As you can check, this is equivalent to your expression.

Another way to establish the relation is to set
$$\lambda(t)=pt+q$$ and solve the system
$$\begin{cases}pc+q=a,\\pd+q=b.\end{cases}$$
