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10.03 Dividing segments

Lesson

Concept summary

We can divide a line segment into two congruent pieces by finding the midpoint. We can also divide a line segment into other proportions.

Endpoint

The point at the end of a segment or the starting point of a ray. A line segment will have two endpoints.

A line segment with a point at either end. One of the points is highlighted.
Midpoint

The point that divides a line segment into two congruent line segments.

A line segment with endpoints A and C with a point B directly in the middle labeled Midpoint

We can find the midpoint of a line segment using the following formula, which shows that the coordinates of the midpoint are the average of the coordinates of the endpoints.

\displaystyle M=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)
\bm{M}
is the midpoint of the line segment
\bm{\left(x_1,y_1\right)}
is the first endpoint of the line segment
\bm{\left(x_2,y_2\right)}
is the second endpoint of the line segment

For endpoints A and B, we may also see this formula written as:M=\frac{1}{2}A+\dfrac{1}{2}B = \frac{A + B}{2}

Notice that this looks the same as taking the mean (average) of two values. To find a mean, we sum the values together and divide by the number of values.

We can extend this idea to finding a weighted average of two or more values. To find a weighted average, we multiply each value by its weight and sum the results, then divide by the total of the weightings. That is,

\displaystyle \text{Weighted Avg} = \frac{w_aA + w_bB + \ldots}{w_a + w_b + \ldots}
\bm{A, B, \ldots}
are the values or quantities to be averaged
\bm{w_a, w_b, \ldots}
are the weightings of A, B, \ldots respectively

If the sum of the weightings is 1, i.e. w_a + w_b + \ldots = 1, then the formula becomes\text{Weighted Avg} = w_aA + w_bB + \ldots

We can directly use this idea to find points at given locations along a line segment.

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Point P is located \dfrac{3}{5} of the way from point A to point B. Taking a weighted average of A and B, with weightings of \dfrac{2}{5} and \dfrac{3}{5} respectively, we have:\begin{aligned} P = &\, \frac{2}{5}A + \frac{3}{5}B\\ = &\, \frac{2}{5}\left(2, 1\right) + \frac{3}{5}\left(7, 11\right)\\ = &\, \left(\frac{2}{5} \cdot 2 + \frac{3}{5} \cdot 7, \enspace \frac{2}{5} \cdot 1 + \frac{3}{5} \cdot 11\right)\\ = &\, \left(5, 7\right) \end{aligned}We can say that P divides \overline{AB} in the ratio 3:2.

In general, we can divide a segment \overline{AB} in the ratio m:n by taking the weighted average\frac{n}{m + n}A + \frac{m}{m + n}BNotice that if m > n the point will be closer to B, and so B has a higher weighting. The reverse is true if m < n.

If the ratio is 1:1, then the resulting point will be the midpoint, and the weighted average becomes the same as the midpoint formula.

Another strategy for partioning a line is to use similar triangles:

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We can count squares or use absolute values to find the lengths of the legs, rise and run, of the right triangle with hypotenuse \overline{AB}

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We can create a similar triangle that is \dfrac{3}{5} the size of the original triangle.

\dfrac{3}{5}\text{run}=\dfrac{3}{5}\cdot 5=3

\dfrac{3}{5}\text{rise}=\dfrac{3}{5}\cdot 10=6

From point A\left(2,1\right), moving right 3 and up 6, we get:

\left(2+3,1+6\right)=\left(5,7\right)

Worked examples

Example 1

Consider the line segment with endpoints A \left(-5,-4\right) and B \left(1,8\right).

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a

Find the midpoint, M, of \overline{AB}.

Solution

\displaystyle M\displaystyle =\displaystyle \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)Formula for midpoint
\displaystyle =\displaystyle \left(\dfrac{-5+1}{2}, \dfrac{-4+8}{2} \right)Subtitute in coordinates of endpoints
\displaystyle =\displaystyle \left(\dfrac{-4}{2}, \dfrac{4}{2} \right)Simplify numerators
\displaystyle =\displaystyle \left(-2, 2 \right)Simplify fractions

Reflection

When finding the midpoint of a line segment, the order of the endpoints doesn't matter, since M is the same distance from both endpoints.

b

Using the weighted average, find the point, P, that is \dfrac{1}{3} of the way from A to B.

Approach

Points that divide the line segment are like finding the weighted average of the two endpoints. In this case, P will divide \overline{AB} in the proportion 2:1.

Solution

\displaystyle P\displaystyle =\displaystyle \dfrac{2}{3}A+\dfrac{1}{3}BWeighted average of the endpoints
\displaystyle =\displaystyle \dfrac{2}{3}\left(-5,-4\right)+\dfrac{1}{3}\left(1,8\right)Substitute in the endpoints
\displaystyle =\displaystyle \left(\dfrac{-10}{3},\dfrac{-8}{3}\right)+\left(\dfrac{1}{3},\dfrac{8}{3}\right)Distribute the multiplication
\displaystyle =\displaystyle \left(\dfrac{-10}{3}+\dfrac{1}{3},\dfrac{-8}{3}+\dfrac{8}{3}\right)Add the corresponding coordinates
\displaystyle =\displaystyle \left(-3,0\right)Simplify

Reflection

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A rough sketch can help us to check our answer.

P does appear to be \dfrac{1}{3} of the way from A to B.

Notice this is the same as being \dfrac{2}{3} of the way from B to A.

Example 2

If the midpoints of A \left(a,b\right) and B \left(1,4\right) is M\left(9,7\right):

a

Find the value of a.

Approach

We can solve for the missing endpoint by substituting the midpoint and known endpoint into the formula for midpoint. To find a, we only need to look at the x-coordinate.

Solution

\displaystyle M\left(x_M, y_M\right)\displaystyle =\displaystyle \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)Formula for midpoint
\displaystyle x_M\displaystyle =\displaystyle \dfrac{x_1+x_2}{2}Only the x-coordinate
\displaystyle 9\displaystyle =\displaystyle \dfrac{a+1}{2}Substitute in the x-coordinates
\displaystyle 18\displaystyle =\displaystyle a+1Multiply both sides by 2
\displaystyle 17\displaystyle =\displaystyle aSubtract 1 from both sides
\displaystyle a\displaystyle =\displaystyle 17Symmetric property of equality
b

Find the value of b.

Approach

To find b, we only need to look at the y-coordinate.

Solution

\displaystyle M\left(x_M, y_M\right)\displaystyle =\displaystyle \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)Formula for midpoint
\displaystyle y_M\displaystyle =\displaystyle \dfrac{y_1+y_2}{2}Only the y-coordinate
\displaystyle 7\displaystyle =\displaystyle \dfrac{b+4}{2}Substitute in the y-coordinates
\displaystyle 14\displaystyle =\displaystyle b+4Multiply both sides by 2
\displaystyle 10\displaystyle =\displaystyle bSubtract 4 from both sides
\displaystyle b\displaystyle =\displaystyle 10Symmetric property of equality

Reflection

This means that the coordinates of A are \left(17,10\right).

Outcomes

MA.912.GR.1.6

Solve mathematical and real-world problems involving congruence or similarity in two-dimensional figures.

MA.912.GR.3.1

Determine the weighted average of two or more points on a line.

MA.912.GR.3.3

Use coordinate geometry to solve mathematical and real-world geometric problems involving lines, circles, triangles and quadrilaterals.

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