8.02 Line segments and distance

Line segments

Remember that lines contain an infinite amount of points, and they are never-ending? Well, sometimes we might want to describe a portion of a line.  We can use line segments to describe these portions.

Suppose we just connect two points with a straight path (no extending onward).  All the points on that path create a line segment.

We can refer to the segment in the diagram above as segment $AB$AB or $\overline{AB}$AB.

The distance between the endpoints of a line segment is called the segment length.  We use different notations to refer to the length of a segment and the segment itself.  For example, the length of $\overline{AB}$AB is denoted $AB$AB and read as "the length of the segment $AB$AB".

Two line segments are congruent when they have the same length.

We place small markings on segments when we want to show that they are congruent. In this diagram, the segment $AB$AB has the same length as the segment $AC$AC.

This does not mean that the two segments are made up of the same points - only that they have the same length. 

Practice question

Question 1

 

Horizontal and vertical distances

Horizontal

Points that lie on a horizontal line segment 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 recognize that the points lie on a horizontal line segment then the distance between them is the distance between the $x$x values: the largest $x$x value minus the smallest $x$x value.

Vertical

Points that lie on a vertical line segment 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 segment could have coordinates $\left(a,b\right)$(a,b) and $\left(a,c\right)$(a,c).

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

Worked example

Question 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$29) minus the smallest $y$y value ($5$5)

$29-5=24$29−5=24

 

The distance formula

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

Well to answer this question we need to return to the Pythagorean theorem.

Exploration

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

 

Sketch them on the Coordinate 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 triangle.

 

 

 

 

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

 

 

 

 

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

 

Recognizing the pattern

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

 

Sketch them on the Coordinate 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 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 $A$A and $B$B)

What we have created here is called the distance formula.

We can use it to find the distance between any $2$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 $A$A and $B$B equal to $5$5? Can you make the distance equal to $1.41$1.41?

 

 

Practice questions

Question 3

Question 4

Betweenness of points

We say that a point $B$B is between two other points $A$A and $C$C if and only if all three points are collinear and $AB+BC=AC$AB+BC=AC.

We can apply the definition of the betweenness of points to solve for unknown lengths in diagrams.

 

Practice questions

Question 5

Question 6