So far we have four different ways to form the equation of a straight line.
We have:
$y=mx+b$y=mx+b (gradient intercept form)
$y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1) (point gradient formula)
$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$y−y1x−x1=y2−y1x2−x1 (two point formula)
$ax+by+c=0$ax+by+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 gradient 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 gradient 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 gradient 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 gradient, we have 4 possible points to choose from and we can use the point gradient formula to find the equation of the line.
$y-y_1=m\left(x-x_1\right)$y−y1=m(x−x1)
Using gradient = $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 gradient 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=y2−y1x2−x1
Gradient 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 |
Gradient 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.
Gradient 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 gradient 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 gradient, $m=-5$m=−5. So we can use the point gradient 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 |