Linear Equations

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.

Relate graphs, tables, and equations to linear, quadratic, and simple exponential relationships found in number and spatial patterns

Investigate relationships between tables, equations and graphs