audio read-through Working with Linear Functions: Finding a New Point, Given a Point and a Slope

Suggested review of lines from the Algebra I curriculum:

Recall:

$$ \begin{align} &\cssId{s7}{\text{slope of line through points } (x_1,y_1) \text{ and } (x_2,y_2)}\cr\cr &\qquad \cssId{s8}{=\ \ m}\cr\cr &\qquad \cssId{s9}{=\ \ \frac{\text{rise}}{\text{run}}}\cr\cr &\qquad\cssId{s10}{=\ \ \frac{\text{change in } y}{\text{change in } x}}\cr\cr &\qquad\cssId{s11}{=\ \ \frac{\Delta y}{\Delta x}}\cr\cr &\qquad\cssId{s12}{=\ \ \frac{y_2 - y_1}{x_2 - x_1}} \end{align} $$

A common scenario is (see diagram below):

finding a new point on a line

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:

$$ \cssId{s39}{\Delta x = 1.4 - 1 = 0.4} $$ $$ \begin{align} \cssId{s40}{y_{\text{new}}}\ &\cssId{s41}{= y_{\text{old}} + m\Delta x}\cr &\cssId{s42}{= -5 + (7.5)(0.4)}\cr &\cssId{s43}{= -2} \end{align} $$

Concept Practice