The point-gradient formula

So far we have two different forms of the equation for a straight line.  

 

What if the information given is a point on the line and the gradient of the line?  

We have a couple of options. 

 

Method 1 - Using $y=mx+b$y=mx+b

We could use this information and construct an equation in gradient intercept form.

Find the equation of a line that passes through the point $\left(2,-8\right)$(2,−8) and has gradient of $-2$−2. 

Think:  We can instantly identify the $m$m value in $y=mx+b$y=mx+b: $m=-2$m=−2

If the point $\left(2,-8\right)$(2,−8) is on the line, then it will satisfy the equation.

Do: $y=mx+b$y=mx+b

$y=-2x+b$y=−2x+b

To find $b$b: we can substitute the values of the point $\left(2,-8\right)$(2,−8)

$-8=-2\times2+b$−8=−2×2+b    and we can now solve for $b$b.  

$-8=-4+b$−8=−4+b

$-4=b$−4=b

So the equation of the line is $y=-2x-4$y=−2x−4

Method 2 - The Point Gradient Formula

Using the same values as the question in Method 1, 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 know that the gradient formula for $m$m is 

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

In this case, our ($x_2$x2​, $y_2$y2​)  is any of the points $\left(x,y\right)$(x,y) and ($x_1$x1​, $y_1$y1​) is the point we are given.

So, 

$m$m	$=$=	$\frac{y_2-y_1}{x_2-x_1}$y2​−y1​x2​−x1​​
$m$m	$=$=	$\frac{y-y_1}{x-x_1}$y−y1​x−x1​​
$m\left(x-x_1\right)$m(x−x1​)	$=$=	$y-y_1$y−y1​
$y-y_1$y−y1​	$=$=	$m\left(x-x_1\right)$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.

 

Example of Method 2

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

$y-y_1$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

A much tidier method than the method used in the previous example!

Further Examples

Let's have a look at some worked solutions.

Question 1

Question 2