4.05 Equation of a straight line

So far we have looked at methods for sketching graphs using the point-gradient form of the equation. We know that all linear equations can be written in the form $y=mx+c$y=mx+c where $m$m is the gradient and $c$c is the value of the $y$y-intercept.

Knowing this, we can also work out the equation of a straight line if we are given its graph (or a table of values) - we just need to work out the gradient and $y$y-intercept. That is, we want to find $m$m and $c$c

To find $c$c we can just look at where the line crosses the $y$y-axis. The value of $y$y at this point is our $y$y-intercept.

To find the gradient, we want to choose two points on the line that we can easily identify the co-ordinates of, ideally points with integer co-ordinates. Using these two points we can calculate the gradient, by identifying the rise and run of the line, or by using the gradient formula.

 

 

The point-gradient formula

Sometimes we don't know the value of the $y$y-intercept, but if we know the gradient and the coordinates of one point, we can still find the equation of the line. If we don't know the gradient, but know the coordinates of two points, we can first find the gradient and then use the point-gradient formula.

 

Exploration

Let's say we know that the gradient of the line is $-2$−2. We also know a point on the line, $\left(2,-8\right)$(2,−8).

Now, apart from this point there are infinitely many other points on this line, and we will let $\left(x,y\right)$(x,y) represent each of them.

Well, since $\left(x,y\right)$(x,y) and $\left(2,-8\right)$(2,−8) are points on the line, then the gradient between them will be $-2$−2.

We know that to find the gradient given two points, we use: 

$m=\frac{y_2-y_1}{x_2-x_1}$m=y2​−y1​x2​−x1​​

Let's apply the gradient formula to $\left(x,y\right)$(x,y) and $\left(2,-8\right)$(2,−8):

$m=\frac{y-\left(-8\right)}{x-2}$m=y−(−8)x−2​

But we know that the gradient of the line is $-2$−2. So:

$\frac{y-\left(-8\right)}{x-2}=-2$y−(−8)x−2​=−2

Rearranging this slightly, we get:

$y-\left(-8\right)=-2\left(x-2\right)$y−(−8)=−2(x−2)

You may be thinking that we should simplify the equation, and of course if you do you should get $y=-2x-4$y=−2x−4 What we want to do though, is generalise our steps so that we can apply it to any case where we're given a gradient $m$m, and a point on the line $\left(x_1,y_1\right)$(x1​,y1​).

In the example above, the point on the line was $\left(2,-8\right)$(2,−8). Let's generalise and replace it with $\left(x_1,y_1\right)$(x1​,y1​).

We were also given the gradient $-2$−2. Let's generalise and replace it with $m$m.

$y-\left(-8\right)=-2\left(x-2\right)$y−(−8)=−2(x−2) becomes $y-y_1=m\left(x-x_1\right)$y−y1​=m(x−x1​)

We call this the point gradient formula, because when we know a point and the gradient using this rule we can easily get the equation of that line.

 

 

Worked example

Find the equation of a line, in point-gradient form, that passes through the point $\left(-4,3\right)$(−4,3) and has gradient of $5$5. 

Think: We know the gradient of the line, and a point that it passes through, so we can use the point-gradient formula to find the equation of the line.

Do: Substitute $m=5$m=5, $x_1=-4$x1​=−4 and $y_1=3$y1​=3 into the equation $y-y_1=m\left(x-x_1\right)$y−y1​=m(x−x1​), and rearrange to solve for $y$y.

$y-y_1$y−y1​	$=$=	$m\left(x-x_1\right)$m(x−x1​)
$y-3$y−3	$=$=	$5\left(x-\left(-4\right)\right)$5(x−(−4))
$y-3$y−3	$=$=	$5\left(x+4\right)$5(x+4)
$y-3$y−3	$=$=	$5x+20$5x+20
$y$y	$=$=	$5x+23$5x+23

We can also achieve the same result by substituting the known values straight into the equation $y=mx+c$y=mx+c. We can then rearrange the equation to solve for the unknown $c$c.

In the above case, we know that $m=5$m=5 and the point $\left(-4,3\right)$(−4,3) lies on the line. Substituting these values into $y=mx+c$y=mx+c gives us:

$3=5\times\left(-4\right)+c$3=5×(−4)+c

Rearranging the equation to solve for $c$c gives us:

$c=23$c=23

We now know the values of $c$c and $m$m and can substitute them into the equation $y=mx+c$y=mx+c, giving us:

$y=5x+23$y=5x+23