﻿ Working with Linear Functions

# WORKING WITH LINEAR FUNCTIONS: FINDING A NEW POINT, GIVEN A POINT AND A SLOPE

by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
• PRACTICE (online exercises and printable worksheets)

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:   $\,\Delta x = 1.4 - 1 = 0.4\,$

$\cssId{s40}{y_{\text{new}}} \cssId{s41}{= y_{\text{old}} + m\Delta x} \cssId{s42}{= -5 + (7.5)(0.4)} \cssId{s43}{= -2}$
Master the ideas from this section