Suggested review of lines from the
Algebra I curriculum:
Recall:
$$
\cssId{s7}{\text{slope of line through points } (x_1,y_1) \text{ and } (x_2,y_2)}
\cssId{s8}{\ \ =\ \ m}
\cssId{s9}{\ \ =\ \ \frac{\text{rise}}{\text{run}}}
\cssId{s10}{\ \ =\ \ \frac{\text{change in } y}{\text{change in } x}}
\cssId{s11}{\ \ =\ \ \frac{\Delta y}{\Delta x}}
\cssId{s12}{\ \ =\ \ \frac{y_2  y_1}{x_2  x_1}}$$
A common scenario is (see diagram at right):
 you have a line with known slope $\,m\,$

you know the coordinates of one point on the line;
call this known point $(x_{\text{old}},y_{\text{old}})$

there is another point on the line whose coordinates are needed;
call this desired point $(x_{\text{new}},y_{\text{new}})$

you know the change in $\,x\,$ between the old and new point:
$$
\cssId{s20}{\Delta x = x_{\text{new}}  x_{\text{old}}}
$$
$$
\begin{alignat}{2}
\cssId{s21}{\Delta x > 0}\,\ \ &
\cssId{s22}{\iff}\ \ &&
\cssId{s23}{\text{the new point lies to the right of the old point}}\cr
\cssId{s24}{\Delta x < 0}\,\ \ &
\cssId{s25}{\iff}\ \ &&
\cssId{s26}{\text{the new point lies to the left of the old point}}
\end{alignat}
$$
 you want the $y$coordinate, $\,y_{\text{new}}\,$, of the new point


Solving for $\,y_{\text{new}}\,$ in terms of the known quantities:
$\displaystyle m = \frac{y_{\text{new}}  y_{\text{old}}}{x_{\text{new}}  x_{\text{old}}}$
$
\cssId{s30}{\Rightarrow}\ \
\cssId{s31}{y_{\text{new}}  y_{\text{old}} = m
\overbrace{(x_{\text{new}}  x_{\text{old}})}^{\Delta x}}$
$
\cssId{s32}{\Rightarrow}\ \
\cssId{s33}{y_{\text{new}} = y_{\text{old}} + m\Delta x}$
EXAMPLE:
You have a known point $\,(1,5)\,$ on a line with slope $\,7.5\,$.
When $\,x = 1.4\,$, what is the $y$value of the point on the line?
SOLUTION:
The change in $\,x\,$ in going from the known point ($x = 1$) to the new point ($x = 1.4$) is:
$\,\Delta x = 1.4  1 = 0.4\,$
$
\cssId{s40}{y_{\text{new}}}
\cssId{s41}{= y_{\text{old}} + m\Delta x}
\cssId{s42}{= 5 + (7.5)(0.4)}
\cssId{s43}{= 2}$