Lesson 1.2 · Unit 1 · G-GPE.6
Adding lengths, and finding the point exactly halfway
The segment addition postulate is the first tool in the course that lets you write an equation from a diagram. On the coordinate plane the same ideas become formulas, and both the distance and midpoint formulas are worth deriving once so that misremembering them is recoverable.
- The segment addition postulate: if \( B \) is between \( A \) and \( C \), then \( AB + BC = AC \).
- Betweenness has to be given or marked. The postulate says nothing about three points until one is known to lie between the others.
- Use it to build an equation from a diagram, substituting the given expressions and solving for the unknown.
- Substitute the solution back to find the lengths asked for, since the question usually wants \( AB \) rather than \( x \).
- The midpoint of a segment divides it into two congruent segments, so \( M \) is the midpoint of \( \overline{AC} \) exactly when \( AM = MC \).
- On the coordinate plane the midpoint is the average of the coordinates: \( M = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) \).
- The distance formula is the Pythagorean theorem in disguise: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), the hypotenuse of a right triangle whose legs are the horizontal and vertical differences.
- Check a midpoint by confirming it is the same distance from each endpoint, and check a segment addition answer by confirming the two parts sum to the whole.
Where students lose marks: subtracting in the midpoint formula. The midpoint averages the coordinates and the distance formula subtracts them. Mixing them up gives an answer that fails the sanity check of lying between the two points.
The problem. (a) Point \( B \) lies between \( A \) and \( C \) with \( AB = 3x + 2 \), \( BC = 2x - 1 \) and \( AC = 26 \). Find \( AB \) and \( BC \). (b) For \( A(-3, 2) \) and \( B(5, 8) \), find the midpoint of \( \overline{AB} \) and the length \( AB \).
Step one: write the postulate for (a) before substituting. Since \( B \) is between \( A \) and \( C \), the segment addition postulate gives \[ AB + BC = AC \] Naming the postulate first is the habit this course is built on: the equation exists because a postulate licenses it, not because the picture suggests it.
Step two: substitute and solve. \( (3x + 2) + (2x - 1) = 26 \), so \( 5x + 1 = 26 \), giving \( 5x = 25 \) and \( x = 5 \).
Step three: answer what was asked. The question wants lengths, not \( x \). \( AB = 3(5) + 2 = 17 \) and \( BC = 2(5) - 1 = 9 \).
Step four: check (a) against the postulate. \( 17 + 9 = 26 \), which is \( AC \). Correct. Both lengths are positive, as lengths must be, which is a second check worth making when an expression could have gone negative.
Step five: find the midpoint in (b) by averaging. \[ M = \left( \frac{-3 + 5}{2}, \frac{2 + 8}{2} \right) = \left( \frac{2}{2}, \frac{10}{2} \right) = (1, 5) \]
Step six: sanity check the midpoint. Its \( x \) coordinate, 1, lies between \( -3 \) and 5, and its \( y \) coordinate, 5, lies between 2 and 8. A midpoint that fails this test has been computed by subtracting instead of averaging.
Step seven: find the length with the distance formula. The horizontal difference is \( 5 - (-3) = 8 \) and the vertical difference is \( 8 - 2 = 6 \). \[ AB = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \] The differences may be taken in either order, since both are squared and squaring removes the sign.
Step eight: verify the midpoint using the distance formula. From \( A(-3,2) \) to \( M(1,5) \): differences 4 and 3, so \( \sqrt{16 + 9} = \sqrt{25} = 5 \). From \( M(1,5) \) to \( B(5,8) \): differences 4 and 3, so also 5. The two halves are equal and they sum to 10, which is \( AB \). The midpoint is confirmed independently of the formula that produced it.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- If \( B \) is between \( A \) and \( C \), \( AB = 7 \) and \( BC = 11 \), find
\( AC \).
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Segment addition postulate. 18
- Find the midpoint of the segment from \( (2, 6) \) to \( (10, 4) \).
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Average each coordinate. \( (6, 5) \)
- Find the distance between \( (0, 0) \) and \( (3, 4) \).
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\( \sqrt{9 + 16} = 5 \). 5
- \( M \) is the midpoint of \( \overline{AB} \) and \( AM = 9 \). Find \( AB \).
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A midpoint makes two congruent segments. 18
- What postulate lets you write \( AB + BC = AC \)?
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The segment addition postulate, provided \( B \) is between \( A \) and \( C \)
- \( B \) is between \( A \) and \( C \) with \( AB = 4x - 3 \), \( BC = x + 5 \) and
\( AC = 32 \). Find all three lengths.
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By the segment addition postulate, \( (4x - 3) + (x + 5) = 32 \), so \( 5x + 2 = 32 \), giving \( 5x = 30 \) and \( x = 6 \). \( AB = 4(6) - 3 = 21 \) and \( BC = 6 + 5 = 11 \). Check: \( 21 + 11 = 32 \). Correct, and both parts are positive. \( AB = 21 \), \( BC = 11 \), \( AC = 32 \)
- Find the distance between \( (-2, 5) \) and \( (4, -3) \).
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Horizontal difference: \( 4 - (-2) = 6 \). Vertical difference: \( -3 - 5 = -8 \). \( d = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \). The negative vertical difference is squared away, which is why the order of subtraction does not matter. 10
- \( M(3, -1) \) is the midpoint of \( \overline{AB} \) and \( A \) is \( (7, 5) \). Find
\( B \).
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The midpoint formula gives two equations. For the \( x \) coordinates: \( \dfrac{7 + x}{2} = 3 \), so \( 7 + x = 6 \) and \( x = -1 \). For the \( y \) coordinates: \( \dfrac{5 + y}{2} = -1 \), so \( 5 + y = -2 \) and \( y = -7 \). So \( B \) is \( (-1, -7) \). Check by averaging: \( \dfrac{7 + (-1)}{2} = 3 \) and \( \dfrac{5 + (-7)}{2} = -1 \), which is \( M \). Correct. A useful shortcut worth knowing: the midpoint is as far from \( B \) as from \( A \), so from \( A(7,5) \) move to \( M(3,-1) \) by going 4 left and 6 down, then repeat the same move to reach \( (-1, -7) \). \( B(-1, -7) \)
- Explain why the distance formula is the Pythagorean theorem.
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Take two points \( (x_1, y_1) \) and \( (x_2, y_2) \). Draw the horizontal segment from the first to \( (x_2, y_1) \) and the vertical segment from there to the second. Those two segments meet at a right angle, because one is horizontal and the other vertical, so together with the segment joining the original points they form a right triangle. The horizontal leg has length \( |x_2 - x_1| \) and the vertical leg \( |y_2 - y_1| \). The segment joining the two points is the hypotenuse. The Pythagorean theorem gives \( d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \), and taking the positive square root gives the distance formula. The absolute values disappear because each quantity is squared. This is worth knowing rather than memorizing the formula, because a forgotten formula can be rebuilt in ten seconds from a sketch. The horizontal and vertical differences are the legs of a right triangle whose hypotenuse is the distance
- \( B \) is between \( A \) and \( C \). \( AB = 2x + 1 \), \( BC = 3x - 4 \), and
\( AB = BC \). Find \( AC \), and say what this makes \( B \).
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Two facts are given, so use them in order. First, \( AB = BC \) gives \( 2x + 1 = 3x - 4 \), so \( x = 5 \). Then \( AB = 2(5) + 1 = 11 \) and \( BC = 3(5) - 4 = 11 \). Equal, as required. By the segment addition postulate, \( AC = 11 + 11 = 22 \). Since \( B \) lies between \( A \) and \( C \) and divides \( \overline{AC} \) into two congruent segments, \( B \) is the midpoint of \( \overline{AC} \) by the definition of a midpoint. Both conditions in the definition matter and both were checked: a point equidistant from \( A \) and \( C \) but off the line is not the midpoint, and betweenness is what rules that out. \( AC = 22 \), and \( B \) is the midpoint of \( \overline{AC} \)