Length of a Line Segment


The length of a line segment is the distance from one endpoint to the other.  

If we are asked to determine the length of a line segment, we could use a ruler to measure the distance between the endpoints. However, if we are given the endpoints of the line segment as points on a Cartesian plane, we can calculate the length more accurately using a distance formula. 


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Developing the Distance Formula

Let's first consider the length of a line segment with one endpoint located at the origin.

Developing the Distance Formula Continued

Next, let's look at the length of a line segment where neither endpoint is located at the origin.

Developing the Distance Formula Continued

\(\begin{align*} (d_{AB})^2&=(\Delta x)^2+(\Delta y)^2\\ d_{AB}&=\sqrt{(3-0)^2+(4-0)^2}\\ d_{AB}&=\sqrt{3^2+4^2}\\ d_{AB}&=\sqrt{25}\\ d_{AB}&=5 \end{align*}\)

A line segment starts from point A (0, 0), and ends at point B (3, 4).

 

Developing the Distance Formula Continued

The length of a line segment with the endpoints \((x_1,y_1)\) and \((x_2,y_2)\) can be calculated using the following distance formula:

 

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Example 5

Determine the exact length of a line segment with endpoints \(A(-10,16)\) and \(B(18,25)\).

Solution

Using the distance formula, we can calculate the length of the line segment.

\(\begin{align*} d_{AB}&=\sqrt{(\Delta x)^2+(\Delta y)^2}\\ d_{AB}&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d_{AB}&=\sqrt{(18-(-10))^2+(25-16)^2}\\ d_{AB}&=\sqrt{(28)^2+(9)^2}\\ d_{AB}&=\sqrt{784+81}\\ d_{AB}&=\sqrt{865} \end{align*}\)

 Therefore, the length of the line segment is exactly \(\sqrt{865}\) units.


Check Your Understanding 3


Determine the exact length of a line segment with endpoints \(A((x)*1.0, (y)*1.0)\) and \(B((x)*2.0, (y)*2.0)\).

Enter \(\sqrt{a}\) as "sqrt\((a)\)".

The length is There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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Try This Revisited

A triangle has vertices \(A(-3,1)\), \(B(2,4)\), and \(C(2,-2)\). Calculate the length of each side to determine if the triangle can be classified as scalene, isosceles, or equilateral.

Solution

Recall the different types of triangles based on their side lengths.

Equilateral

All three side lengths
are the same.

A yield sign has the shape of an equilateral triangle.

Source: Yield - Brilt/iStock/Thinkstock

Isosceles

Two side lengths are
the same.

The roof of a house takes the shape of an isosceles triangle.

Source: House - rickmartinez/iStock/Thinkstock

Scalene

All three sides have
different lengths.

A boat sail has the shape of a scalene triangle.

Source: Sailboat - Con Tanasiuk / Design Pics/ Valueline/Thinkstock

Calculate the length of \(AB\)

\(A(-3,1)\) and \(B(2,4)\)

\(\begin{align*} d_{AB}&=\sqrt{(\Delta x)^2+(\Delta y)^2}\\ d_{AB}&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d_{AB}&=\sqrt{(2-(-3))^2+(4-1)^2}\\ d_{AB}&=\sqrt{(5)^2+(3)^2}\\ d_{AB}&=\sqrt{25+9}\\ d_{AB}&=\sqrt{34} \end{align*}\)

Calculate the length of \(BC\)

\(B(2,4)\) and \(C(2,-2)\)

\(\begin{align*} d_{BC}&=\sqrt{(\Delta x)^2+(\Delta y)^2}\\ d_{BC}&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d_{BC}&=\sqrt{(2-2)^2+(-2-4)^2}\\ d_{BC}&=\sqrt{(0)^2+(-6)^2}\\ d_{BC}&=\sqrt{0+36}\\ d_{BC}&=\sqrt{36}\\ d_{BC}&=6 \end{align*}\)

Calculate the length of \(AC\)

\(A(-3,1)\) and \(C(2,-2)\)

\(\begin{align*} d_{AC}&=\sqrt{(\Delta x)^2+(\Delta y)^2}\\ d_{AC}&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ d_{AC}&=\sqrt{(2-(-3))^2+(-2-1)^2}\\ d_{AC}&=\sqrt{(5)^2+(-3)^2}\\ d_{AC}&=\sqrt{25+9}\\ d_{AC}&=\sqrt{34} \end{align*}\)

Therefore, because \(d_{AB}=d_{AC}\ne d_{BC}\), the triangle is isosceles.

Let's look at one last example where we can use both the midpoint and length of a line segment to answer the question.


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Example 6

Determine the midpoint of the line segment with endpoints \(A(-8,6)\) and \(B(12,10)\).

Verify that this is the midpoint by calculating the distance from the midpoint to each endpoint on the line segment.

 

 

 

Example 6 Continued

Determine the midpoint of the line segment with endpoints \(A(-8,6)\) and \(B(12,10)\).

Verify this is the midpoint by calculating the distance from the midpoint to each endpoint on the line segment.

Solution

Using the calculated midpoint \(P(2,8)\):

\(\begin{align*} \class{timed cover remove1-cover}{d_{AP}}&\class{timed cover remove1-cover}{=\sqrt{(\Delta x)^2+(\Delta y)^2}}\\ &\class{timed cover remove1-cover}{=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\ &\class{timed cover remove2-cover}{=\sqrt{(-8-2)^2+(6-8)^2}}\\ &\class{timed cover remove3-cover}{=\sqrt{(-10)^2+(-2)^2}}\\ &\class{timed cover remove3-cover}{=\sqrt{100+4}}\\ &\class{timed cover remove4-cover}{=\sqrt{104}} \end{align*}\)

\(\begin{align*} \class{timed cover remove5-cover}{d_{PB}}&\class{timed cover remove5-cover}{=\sqrt{(\Delta x)^2+(\Delta y)^2}}\\ &\class{timed cover remove5-cover}{=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\ &\class{timed cover remove6-cover}{=\sqrt{(2-12)^2+(8-10)^2}}\\ &\class{timed cover remove7-cover}{=\sqrt{(-10)^2+(-2)^2}}\\ &\class{timed cover remove7-cover}{=\sqrt{100+4}}\\ &\class{timed cover remove8-cover}{=\sqrt{104}} \end{align*}\)

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