Linear Equations
UK Secondary (7-11)
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The point-gradient formula
Lesson

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

Equation of Lines!

We have:

$y=mx+b$y=mx+b  (gradient intercept form)

$ax+by+c=0$ax+by+c=0   (general form)

 

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=2x4

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=y2y1x2x1

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)x2

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

$\frac{y-\left(-8\right)}{x-2}=-2$y(8)x2=2

Rearranging this slightly, we get:

$y-\left(-8\right)=-2\left(x-2\right)$y(8)=2(x2)

You may be thinking that we should simplify the equation, and of course if you do you should get $y=-2x-4$y=2x4 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(x2)      becomes     $y-y_1=m\left(x-x_1\right)$yy1=m(xx1)

 

The Point-Gradient Formula

Given a point on the line $\left(x_1,y_1\right)$(x1,y1) and the gradient $m$m, the equation of the line is:

$y-y_1=m\left(x-x_1\right)$yy1=m(xx1)

 

We know that the gradient formula for $m$m is 

$m=\frac{y_2-y_1}{x_2-x_1}$m=y2y1x2x1

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 $=$= $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1
$m$m $=$= $\frac{y-y_1}{x-x_1}$yy1xx1
$m\left(x-x_1\right)$m(xx1) $=$= $y-y_1$yy1
$y-y_1$yy1  $=$= $m\left(x-x_1\right)$m(xx1)

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

$y-y_1$yy1 $=$= $m\left(x-x_1\right)$m(xx1)
$y-3$y3 $=$= $5\left(x-\left(-4\right)\right)$5(x(4))
$y-3$y3 $=$= $5\left(x+4\right)$5(x+4)
$y-3$y3 $=$= $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

A line passes through Point $A$A $\left(8,2\right)$(8,2) and has a gradient of $2$2.

  1. Find the value of the $y$y-intercept of the line, denoted by $b$b.

  2. Hence, write the equation of the line in gradient-intercept form.

Question 2

A line passes through the point $A$A$\left(-4,3\right)$(4,3) and has a gradient of $-9$9. Using the point-gradient formula, express the equation of the line in gradient intercept form.

 

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