Linear Relations

Ontario 09 Applied (MFM1P)

Finding the rule II

Lesson

So far we have four different ways to form the equation of a straight line.

Equations of Lines!

We have:

$y=mx+b$`y`=`m``x`+`b` (slope intercept form)

$y-y_1=m\left(x-x_1\right)$`y`−`y`1=`m`(`x`−`x`1) (point slope formula)

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$`y`−`y`1`x`−`x`1=`y`2−`y`1`x`2−`x`1 (two point formula)

$ax+by+c=0$`a``x`+`b``y`+`c`=0 (general form)

**Question**: If we are given a table of values, how can we be sure it forms a straight line? And how can we find the equation of that line?

We can't just assume the points in a table of values are linear (make a straight line). So when given a table of values, we must first confirm that the values in the table form a linear relationship.

You can do this a couple of ways, depending on what sort of information is provided to you in the table.

What is it that makes a straight line? It is the fact that for each 1 unit change in $x$`x`, the $y$`y` value changes by the same amount each time (i.e. the slope is always the same).

If the table shows you values of $y$`y`, for consecutive values of $x$`x` (like this one) then it is quite easy to both check for a linear relationship and find the slope of that relationship.

Some things to note about this table: the $x$`x` values go up by 1 each time and it doesn't matter what number the table starts at.

Notice in the above table, that for each 1 unit increase in $x$`x`, $y$`y` increases by $3$3 each time. This is a linear relationship, and what's more, this constant change in $y$`y` is the * slope* of the line.

So we can check for a linear relationship by looking for a common difference between the $y$`y` values. If each successive $y$`y` value has the same difference then it is linear.

From here we not only have the slope, we have 4 possible points to choose from and we can use the point slope formula to find the equation of the line.

$y-y_1=m\left(x-x_1\right)$`y`−`y`1=`m`(`x`−`x`1)

Using slope = $3$3 and the point $\left(3,12\right)$(3,12):

$y-12$y−12 |
$=$= | $3\left(x-3\right)$3(x−3) |

$y-12$y−12 |
$=$= | $3x-9$3x−9 |

$y$y |
$=$= | $3x-9+12$3x−9+12 |

$y$y |
$=$= | $3x+3$3x+3 |

If the table shows you values of $y$`y`, for non-consecutive values of $x$`x` (like this one) then we need another way to check for a linear relationship.

Some things to note about this table: the $x$`x` values do not go up by 1 each time.

We can check for a linear relationship by either plotting accurately, and looking to see if it is a straight line, or by checking the slope between all the points. We will need to do this 3 times in this table.

$m=\frac{y_2-y_1}{x_2-x_1}$`m`=`y`2−`y`1`x`2−`x`1

Slope between $\left(-3,25\right)$(−3,25) and $\left(2,0\right)$(2,0):

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$m$m |
$=$= | $\frac{0-25}{2-\left(-3\right)}$0−252−(−3) |

$m$m |
$=$= | $\frac{-25}{5}$−255 |

$m$m |
$=$= | $-5$−5 |

Slope between $\left(2,0\right)$(2,0) and $\left(10,-40\right)$(10,−40):

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$m$m |
$=$= | $\frac{-40-0}{10-2}$−40−010−2 |

$m$m |
$=$= | $\frac{-40}{8}$−408 |

$m$m |
$=$= | $-5$−5 |

Looking good so far. We now have shown that the $3$3 points $\left(2,0\right)$(2,0), $\left(10,-40\right)$(10,−40) and $\left(-3,25\right)$(−3,25) are collinear which means they are all on the one line. We have to confirm that the last point is also on this line.

Slope between $\left(2,0\right)$(2,0) and $\left(12,-50\right)$(12,−50):

(see how I used the $\left(2,0\right)$(2,0) point again, this is because 0 values make evaluating the slope easier)

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$m$m |
$=$= | $\frac{-50-0}{12-2}$−50−012−2 |

$m$m |
$=$= | $\frac{-50}{10}$−5010 |

$m$m |
$=$= | $-5$−5 |

Got it! A linear relationship.

Now we can find its equation. We already have the slope, $m=-5$`m`=−5. So we can use the point slope formula.

$y-y_1$y−y1 |
$=$= | $m\left(x-x_1\right)$m(x−x1) |

$y-0$y−0 |
$=$= | $-5\left(x-2\right)$−5(x−2) |

$y$y |
$=$= | $-5x+10$−5x+10 |

Let's have a look at these worked examples.

Use the table of values below to write an equation for $g$`g` in terms of $f$`f`.

$f$f |
$4$4 | $5$5 | $6$6 | $7$7 | $8$8 |
---|---|---|---|---|---|

$g$g |
$8$8 | $10$10 | $12$12 | $14$14 | $16$16 |

Determine values of a linear relation by using a table of values, by using the equation of the relation, and by interpolating or extrapolating from the graph of the relation