# The point-slope 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  (slope 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 slope 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 slope intercept form.

Find the equation of a line that passes through the point $\left(2,-8\right)$(2,8) and has slope 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 Slope Formula

Using the same values as the question in Method 1, we know that the slope 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 slope between them will be $-2$2.

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

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

Let's apply the slope 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 slope 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 slope $m$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 slope $-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-Slope Formula

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

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

We know that the slope 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 $=$= $\frac{y_2-y_1}{x_2-x_1}$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 slope formula, because when we know a point and the slope 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 slope 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

A line passes through Point $A$A $\left(8,2\right)$(8,2) and has a slope 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 slope-intercept form.

##### Question 2

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

### Outcomes

#### 10P.LR2.06

Determine the equation of a line, given its graph, the slope and y-intercept, the slope and a point on the line, or two points on the line.