Linear Equations

UK Secondary (7-11)

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`=`m``x`+`b` (gradient intercept form)

$ax+by+c=0$`a``x`+`b``y`+`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.

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`=`m``x`+`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`=`m``x`+`b`

$y=-2x+b$`y`=−2`x`+`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`=−2`x`−4

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`=`y`2−`y`1`x`2−`x`1

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`=−2`x`−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 $\left(x_1,y_1\right)$(`x`1,`y`1).

In the example above, the point on the line was $\left(2,-8\right)$(2,−8). Let's generalise and replace it with $\left(x_1,y_1\right)$(`x`1,`y`1). 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`−`y`1=`m`(`x`−`x`1)

The Point-Gradient Formula

Given a point on the line $\left(x_1,y_1\right)$(`x`1,`y`1) and the gradient $m$`m`, the equation of the line is:

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

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

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

In this case, our ($x_2$`x`2, $y_2$`y`2) is any of the points $\left(x,y\right)$(`x`,`y`) and ($x_1$`x`1, $y_1$`y`1) is the point we are given.

So,

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

$m$m |
$=$= | $\frac{y-y_1}{x-x_1}$y−y1x−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.

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!

Let's have a look at some worked solutions.

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

**Find the value of the $y$**`y`-intercept of the line, denoted by $b$`b`.**Hence, write the equation of the line in gradient-intercept form.**

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.