The Distance Formula

Horizontal and Vertical Distances

Horizontal

Points that lie on a horizontal line share the same $y$y value.  They would have coordinates that look something like this: $\left(2,5\right)$(2,5) and $\left(-4,5\right)$(−4,5). More generally, two points that lie on a horizontal line could have coordinates $\left(a,b\right)$(a,b) and $\left(c,b\right)$(c,b).

If you can recognise that the points lie on a horizontal line then the distance between them is the distance between the $x$x values: the largest $x$x value minus the smallest $x$x value.

Example 1

Find the distance between $\left(2,5\right)$(2,5) and $\left(-4,5\right)$(−4,5).

Think:  Notice that because the $y$y values are the same, these points lie on a horizontal line.

Do: The distance between them will be the largest $x$x value (2) minus the smallest $x$x value (-4)

$2-\left(-4\right)=2+4$2−(−4)=2+4 = $6$6 (remember with adjacent signs, two negatives become addition).  

 

Vertical

Points that lie on a vertical line share the same $x$x value.  They would have coordinates that look something like this: $\left(2,5\right)$(2,5) and $\left(2,29\right)$(2,29). More generally, two points that lie on a vertical line could have coordinates $\left(a,b\right)$(a,b) and $\left(a,c\right)$(a,c).

If you can recognise that the points lie on a vertical line then the distance between them is the distance between the $y$y values: the largest $y$y value minus the smallest $y$y value.

 

Example 2

Find the distance between $\left(2,5\right)$(2,5) and $\left(2,29\right)$(2,29).

Think:  Notice that because the $x$x values are the same, these points lie on a vertical line.

Do: The distance between them will be the largest $y$y value (29) minus the smallest $y$y value (5)

$29-5=24$29−5=24 

Distance Formula

What if the points are not in a horizontal or vertical line?  How can we find the distance then?

Well to answer this question we need to return to Pythagoras' Theorem.  

Example with numbers

Let's have a look at these two points:  A$\left(5,4\right)$(5,4) and B$\left(-3,6\right)$(−3,6).

 

Sketch them on the Cartesian plane.

 

 

 

 

 

 

 

Draw a right triangle connecting the two points, like this.

 

 

 

 

 

On the diagram mark the horizontal and vertical distances (calculate them like we did above). These are now two of the side lengths of the right-angled triangle.

 

 

 

 

Use Pythagoras's theorem to calculate the distance on the hypotenuse.

 

 

 

 

So the distance between A$\left(5,4\right)$(5,4) and B$\left(-3,6\right)$(−3,6) is 8.25 units (to 2 decimal places). 

 

Example with letters

Let's do the same thing now, but with any two general points A($x_1$x1​, $y_1$y1​) and B($x_2$x2​, $y_2$y2​)

 

Sketch them on the Cartesian plane. (We don't really know where these points are so we can represent them like this).

 

 

 

 

 

Draw a right triangle connecting the two points like this.

 

 

 

 

 

On the diagram mark the horizontal and vertical distances (calculate them like we did above). These are two of the side lengths of the right-angled triangle.

 

 

 

Use Pythagoras's theorem to calculate the distance on the hypotenuse.

$c^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$c2=(x2​−x1​)2+(y2​−y1​)2

$c=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$c=√(x2​−x1​)2+(y2​−y1​)2

(where $c$c is the distance between A and B)

What we have created here is called the distance formula.

We can use it to find the distance between ANY 2 points on the plane.

This applet will let you see the distance formula in action.  Move the blue points around and watch the values in the formula change.  Can you make the distance between A and B equal to 5?  Can you make the distance equal to 1.41?

Here are some worked examples.

Question 1

Question 2