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10. Connecting Algebra and Geometry through Coordinates
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Geometry

10.02 Midpoints

Lesson
Practice

Introduction

In this lesson, we will find midpoints using formulas and similar triangles.

Midpoints

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

Exploration

Three line segments are graphed below.

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  1. Find the midpoint of each of the segments.
  2. Describe your strategy for finding the midpoint of \left(2,3\right) and \left(6,11\right).
  3. Think of and describe a second strategy for finding the midpoint of \left(2,3\right) and \left(6,11\right).
  4. How are the two strategies related?

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 \left(x_M,y_M\right)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)
\bm{\left(x_M,y_M\right)}
coordinates of the midpoint
\bm{\left(x_1,y_1\right)}
coordinates of the first endpoint
\bm{\left(x_2,y_2\right)}
coordinates of the second endpoint

Another strategy for finding the midpoint is to use similar triangles. We can count squares or use absolute values to find the lengths of the legs, the rise and run, of the right triangle with hypotenuse \overline{AB}. Then, we can create a similar triangle that is \frac{1}{2} the size of the original triangle.

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Original triangle with hypontenuse \overline{AB}
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New triangle with hypontenuse \frac{1}{2}\left(\overline{AB}\right)

Examples

Example 1

Consider the line segment with endpoints A \left(-5,12\right) and B \left(10,-4\right).

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

Worked Solution
Create a strategy

For this problem, we will use similar triangles to solve. Using this method, we create a triangle with \overline{AB} as the hypotenuse, find the lengths of the legs of the triangle, then create a similar triangle that is half the size of the original.

Apply the idea
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Using the absolute value to find the lengths of the legs, we get:

\text{Leg } 1=\left|-5-10\right|=15

\text{Leg }2=\left|12-\left(-4\right)\right|=16

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The legs of the new triangle need to be half the size of the legs of the original triangle:

\text{New run}=\frac{1}{2}\cdot 15=7.5

\text{New rise}=\frac{1}{2}\cdot 16=8

From point A\left(-5,12\right) moving right 7.5 units and down 8 units, we get the midpoint:

M=\left(-5+7.5, 12-8\right)=\left(2.5,4\right)

Reflect and check

We could have used the midpoint formula to find the answer instead, but we can also use it to check our answer.

\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+10}{2}, \dfrac{12+-4}{2} \right)Substitute coordinates of the endpoints
\displaystyle =\displaystyle \left(\dfrac{5}{2}, \dfrac{8}{2} \right)Evaluate the numerators
\displaystyle =\displaystyle \left(2.5, 4 \right)Evaluate the division

Example 2

The midpoint of A \left(1,4\right) and B \left(a,b\right) is M\left(9,7\right).

a

Find the value of a.

Worked Solution
Create a strategy

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.

Apply the idea
\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{1+a}{2}Substitute the x-coordinates
\displaystyle 18\displaystyle =\displaystyle 1+aMultiply 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.

Worked Solution
Create a strategy

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

Apply the idea
\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{4+b}{2}Substitute the y-coordinates
\displaystyle 14\displaystyle =\displaystyle 4+bMultiply both sides by 2
\displaystyle 10\displaystyle =\displaystyle bSubtract 4 from both sides
\displaystyle b\displaystyle =\displaystyle 10Symmetric property of equality
Reflect and check

This means that the coordinates of B are \left(17,10\right). We can check that this answer is correct by graphing \overline{AB} and using similar triangles.

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The values of the rise and run of the triangle whose hypotenuse is \overline{AB} are 6 and 16 respectively.

\dfrac{1}{2}\cdot \text{run}=\dfrac{1}{2}\cdot 16=8

\dfrac{1}{2}\cdot \text{rise}=\dfrac{1}{2}\cdot 6=3

From point A\left(1,4\right), moving right 8 and up 3, we get:

\left(1+8,4+3\right)=\left(9,7\right)

This shows the given midpoint is halfway between A and B which means our coordinates for B were correct.

Example 3

There is a stretch of nearly perfectly straight road between Dalhart, Texas and Liberal, Kansas along US-54 W.

Dalhart, Texas is located at coordinates of \left(36.06 \degree \text{N}, -102.52 \degree \text{W}\right) and Liberal, Kansas is located at \left(37.05 \degree \text{N}, -100.92 \degree \text{W}\right). Seven cities on this route are given with their coordinates. State which city is closest to the midpoint between Dalhart and Liberal.

  • Conlen, \left(36.24 \degree \text{N}, -102.23 \degree \text{W}\right)
  • Stratford, \left(36.34 \degree \text{N}, -102.07 \degree \text{W}\right)
  • Texhoma, \left(36.51 \degree \text{N}, -101.78 \degree \text{W}\right)
  • Goodwell, \left(36.60 \degree \text{N}, -101.64 \degree \text{W}\right)
  • Guymon, \left(35.68 \degree \text{N}, -101.48 \degree \text{W}\right)
  • Optima, \left(36.76 \degree \text{N}, -101.35 \degree \text{W}\right)
  • Tyrone, \left(36.96 \degree \text{N}, -101.07 \degree \text{W}\right)
Worked Solution
Create a strategy

We can find the midpoint algebraically and then check the map to see which city is closest and ensure we get a reasonable answer.

Apply the idea

Finding the midpoint between: Dalhart, Texas\left(36.06 \degree \text{N}, -102.52 \degree \text{W}\right) and Liberal, Kansas\left(37.05 \degree \text{N}, -100.92 \degree \text{W}\right)

\displaystyle M\displaystyle =\displaystyle \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)Formula for midpoint
\displaystyle M\displaystyle =\displaystyle \left(\dfrac{36.06+37.05}{2}, \dfrac{-102.52+(-100.92)}{2}\right)Substitute the coordinates
\displaystyle M\displaystyle =\displaystyle \left(36.555, -101.72\right)Evaluate the addition and division

The two cities that are close to this midpoint are:

  • Texhoma, \left(36.51 \degree \text{N}, -101.78 \degree \text{W}\right)
  • Goodwell, \left(36.60 \degree \text{N}, -101.64 \degree \text{W}\right)

The x-coordinates are both 0.045 \degree away, but for Texhoma, the y-coordinate is 0.6 \degree away, while Goodwell is 0.8 \degree away. This means that Texhoma is closer to the halfway mark between Dalhart and Liberal.

Reflect and check

Using the map, we could estimate that it would have been one of Texhoma and Goodwell, but we need to work algebraically to confirm which is closer.

Idea summary

We can find the midpoint of a line segment using the midpoint formula:

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

or using similar triangles. To create a triangle that is half the size of the original, we find:

\left(\frac{1}{2}\right)\cdot\text{rise}\\ \left(\frac{1}{2}\right)\cdot\text{run}

and add those values to the coordinates of the leftmost point.

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