2.20 Point-slope form

Guiding questions

As you explore the applet and read the lesson below, think about your answers to the following questions.  Discuss your answers with a partner after reading the lesson.

1. If we are given a point on the line and the slope of the line, what is the best way to find the equation of the line? 
2. In what ways are the slope formula and the point-slope form connected to one another?

 

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

If we are given a point on the line and the slope of the line, what is the best way to find the equation 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.

Worked example

Find the equation of the 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, so we can substitute it in to find the value of $b$b.

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

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

 

Method 2 - The point-slope formula

We are going to derive a new equation for a line using what we know about the equation for slope.

Worked example

Let's do the same question as we did for Method 1, so we want to find the equation of the line that passes through the point $\left(2,-8\right)$(2,−8) and has slope of $-2$−2. 

Think: 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=y2​−y1​x2​−x1​​

Do: 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)x−2​

But we know that the slope 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 generalize 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 generalize and replace it with $\left(x_1,y_1\right)$(x1​,y1​). We were also given the slope $-2$−2. Let's generalize 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​)

 

 

Worked example

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​)	Use the point-slope formula
$y-3$y−3	$=$=	$5\left(x-\left(-4\right)\right)$5(x−(−4))	Fill in the given values
$y-3$y−3	$=$=	$5\left(x+4\right)$5(x+4)	Simplify
$y-3$y−3	$=$=	$5x+20$5x+20
$y$y	$=$=	$5x+23$5x+23	Re-arrange to slope-intercept form

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

 

Practice questions

question 1

question 2

Question 3