3.05 Find and use the equation of a line

 

 

In the line $y=mx+c$y=mx+c the values of $m$m and $c$c have specific meanings. Confirm for yourself what these values do by exploring with this interactive.

 

The gradient

What you will have found is that the value of $m$m affects the gradient.

• If $m<0$m<0, the gradient is negative and the line is decreasing
• if $m>0$m>0, the gradient is positive and the line is increasing
• if $m=0$m=0 the gradient is $0$0 and the line is horizontal
• The larger the magnitude of $m$m the steeper the line

The $y$y-intercept

We also found that the value of $c$c affects the $y$y-intercept.  

• If $c$c is positive then the line cross the $y$y-axis above the origin.
• If $c$c is negative then the line cross the $y$y-axis below the origin.

 

Worked examples

EXAMPLE 1

Consider the linear equation $y=3x$y=3x. 

(a) What is the the gradient?  

Think: For all equations of the form $y=mx+c$y=mx+c, the gradient is the value of $m$m, the coefficient of $x$x.

Do: We can see from the equation that the coefficient of $x$x is $3$3. This means the gradient of this line is $3$3.

(b) What is the $y$y-intercept?  

Think: The $y$y-intercept is the value of $c$c, the constant term.

Do: We can see from the equation that there is no value of $c$c. This is the same as if the equation were $y=3x+0$y=3x+0. This means the $y$y-intercept is $0$0, that is the line crosses the $y$y-axis at the point $\left(0,0\right)$(0,0).

Reflect: All lines of the form $y=mx$y=mx, that is with a $y$y-intercept of $0$0, pass through the origin. This line has a gradient of $3$3, which means it is $3$3 times steeper than the line $y=x$y=x.

 

example 2

Consider the linear equation $y=\frac{x}{2}-3$y=x2​−3.

(a) What is the the gradient?  

Think: For all equations of the form $y=mx+c$y=mx+c, the gradient is the value of $m$m, the coefficient of $x$x.

Do: We can see from the equation that the coefficient of $x$x is $\frac{1}{2}$12​, ($\frac{x}{2}$x2​ is equivalent to $\frac{1}{2}x$12​x) This means the gradient of this line is $\frac{1}{2}$12​.

(b) What is the $y$y-intercept?  

Think: The $y$y-intercept is the value of $c$c, the constant term.

Do: We can see from the equation that $c=-3$c=−3. This means the $y$y-intercept is $-3$−3, that is the line crosses the $y$y-axis at the point $\left(0,-3\right)$(0,−3).

Reflect: The line has a gradient of $\frac{1}{2}$12​ which means it will be half as steep as the line $y=x$y=x. The $y$y-intercept is $-3$−3 which means it will cross the $y$y-axis below the origin.

An equation of the form $y=mx+c$y=mx+c is known as the gradient-intercept form of a line, as we can easily identify both the gradient and the value of the $y$y-intercept.

 

 

Finding the equation of a line

Sometimes we are given the graph of a line, and we are asked to find the equation of the line.

The first thing we want to do is find the gradient of the line, which we can do by using any two points (usually the intercepts). Using the coordinates of the two points, we can either use the gradient formula, or by calculating the rise and the run.

We can then identify the $y$y-intercept, by looking where the line crosses the $y$y-axis. Once we have identified these two features, we can write the equation of the line in the form $y=mx+c$y=mx+c.

 

Practice questions

Question 1

Question 2

Question 3

 

 

The point-gradient formula

Sometimes we don't know the value of the $y$y-intercept, but if we know the gradient and the coordinates of one point, we can still find the equation of the line. If we don't know the gradient, but know the coordinates of two points, we can first find the gradient and then use the point-gradient formula.

 

Exploration

Let's say 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=y2​−y1​x2​−x1​​

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=−2x−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 $\left(x_1,y_1\right)$(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(x−2) becomes $y-y_1=m\left(x-x_1\right)$y−y1​=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.

 

 

Worked example

example 3

Find the equation of a line, in point-gradient form, that passes through the point $\left(-4,3\right)$(−4,3) and has gradient of $5$5. 

Think: We know the gradient of the line, and a point that it passes through, so we can use the point-gradient formula to find the equation of the line.

Do: Substitute $m=5$m=5, $x_1=-4$x1​=−4 and $y_1=3$y1​=3 into the equation $y-y_1=m\left(x-x_1\right)$y−y1​=m(x−x1​), and rearrange to solve for $y$y.

$y-y_1$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

We can also achieve the same result by substituting the known values straight into the equation $y=mx+c$y=mx+c. We can then rearrange the equation to solve for the unknown $c$c.

In the above case, we know that $m=5$m=5 and the point $\left(-4,3\right)$(−4,3) lies on the line. Substituting these values into $y=mx+c$y=mx+c gives us:

$3=5\times\left(-4\right)+c$3=5×(−4)+c

Rearranging the equation to solve for $c$c gives us:

$c=23$c=23

We now know the values of $c$c and $m$m and can substitute them into the equation $y=mx+c$y=mx+c, giving us:

$y=5x+23$y=5x+23