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6.6 — Yard Work in Segments
1. In class: A Solidify Understanding Task
Finding the midpoint of a segment
A midpoint splits a line segment into two parts of equal length. On a number line, the number halfway between 4 and 10 is 7. In the coordinate plane, you find the halfway value separately for the horizontal and vertical coordinates.
Let the endpoints be \(A=(-9,6)\) and \(B=(3,10)\). Moving from A to B requires going 12 units right and 4 units up:
\[\Delta x=3-(-9)=12,\qquad \Delta y=10-6=4.\]Half the move is 6 units right and 2 units up. Starting at A gives
\[(-9+6,\ 6+2)=\boxed{(-3,8)}.\]Check from this point to B: the remaining move is also 6 right and 2 up. Therefore the two halves have equal length.
The full instruction in the MVP packet reads:
“2. Explain to Malik why the logic that made him think the midpoint was (6, 2) is almost right, and how to extend his thinking to use the coordinates of the endpoints to get the midpoint of (-3,6).”
The previous paragraph says Malik found (−3, 8), which is correct: the endpoints given in Problem 1 are \((-9,6)\) and \((3,10)\), whose midpoint is \((-3,8)\). The final coordinate in Problem 2 is therefore a typo: “(-3,6)” should read “(-3,8).”
Why the average gives the correct coordinates
If the segment begins at \((x_1,y_1)\) and ends at \((x_2,y_2)\), then the total horizontal move is \(x_2-x_1\) and the total vertical move is \(y_2-y_1\). Halfway from the first point is
\[x_M=x_1+\frac{x_2-x_1}{2}=\frac{x_1+x_2}{2},\] \[y_M=y_1+\frac{y_2-y_1}{2}=\frac{y_1+y_2}{2}.\]For endpoints \((x_1,y_1)\) and \((x_2,y_2)\),
\[\boxed{M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).}\]Average the x-coordinates; average the y-coordinates. Do not average all four numbers together.
Find the midpoint of \((-5,4)\) and \((2,9)\):
\[M=\left(\frac{-5+2}{2},\frac{4+9}{2}\right)=\boxed{\left(-\frac32,\frac{13}{2}\right)}.\]The midpoint need not have integer coordinates.
Connecting to the two congruent right triangles
In the packet’s drawing, Sapana calls the unknown midpoint \((x,y)\). Because the small right triangles have equal horizontal and vertical legs, their horizontal distances satisfy
\[x-(-9)=3-x\quad\Longrightarrow\quad 2x=-6\quad\Longrightarrow\quad x=-3.\]Their vertical distances satisfy
\[y-6=10-y\quad\Longrightarrow\quad 2y=16\quad\Longrightarrow\quad y=8.\]This proves the same result without first assuming the coordinates of the midpoint.
Dividing a segment in a given ratio
A midpoint divides a segment into two equal pieces. A point can also divide a segment into unequal pieces. For example, if \(AP:PB=1:2\), imagine dividing the whole segment into three equal parts: one from A to P and two from P to B. That puts P one-third of the way from A to B.
We already know how to find a midpoint by taking half the horizontal change and half the vertical change. The same idea works here: take one-third of each coordinate change. That keeps P on the straight line between A and B.
Find P on the segment from \(A=(2,1)\) to \(B=(11,7)\) such that \(AP:PB=1:2\).
From A to B, we move \(11-2=9\) units right and \(7-1=6\) units up. There are three equal parts, so one part is a move of \((9/3,\,6/3)=(3,2)\).
Starting at A, make that one move:
\[P=(2+3,\;1+2)=\boxed{(5,3)}.\]Check: A to P is a move of \((3,2)\). P to B is \((6,4)\)—twice the same move—so the segment lengths are in the required ratio \(1:2\).
Find P between \(A=(4,10)\) and \(B=(-6,-5)\) so that \(AP:PB=2:3\).
There are five equal parts in all. From A to B, the horizontal change is \(-6-4=-10\) and the vertical change is \(-5-10=-15\). One part is therefore a move of \((-2,-3)\).
To reach P, take two of those parts, moving \((-4,-6)\) from A:
\[P=(4-4,\;10-6)=\boxed{(0,4)}.\]Check: P to B is a move of \((-6,-9)\), exactly three of the same equal parts. The two pieces have lengths in the ratio \(2:3\).
In general: add the numbers in the ratio to find the total number of equal parts. Work out the horizontal and vertical changes from the first endpoint to the second. Divide each change by the total number of parts, then move the required number of parts from the first endpoint. The steps are the same whether the coordinates increase or decrease.
2. Ready review — Arithmetic means and halfway values
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The arithmetic mean is the sum divided by the number of values. The average of two numbers gives the point exactly halfway between them on a number line.
The mean of \(6,10,14,18\) is
\[\frac{6+10+14+18}{4}=\frac{48}{4}=12.\]The mean lies between the smallest and largest values. It need not be one of the listed numbers.
The halfway value between \(-17\) and \(-5\) is
\[\frac{-17+(-5)}{2}=\frac{-22}{2}=\boxed{-11}.\]Check: \(-11-(-17)=6\) and \(-5-(-11)=6\).
The halfway value between \(-18\) and \(7\) is
\[\frac{-18+7}{2}=\boxed{-\frac{11}{2}}.\]Do not replace negative numbers with their absolute values before averaging.
Check your work — Ready 1–12 answer key
- Mean \(9\), between \(6\) and \(12\); deviations \(-3,3,1,-1\) total zero.
- Mean \(7\), the middle value and halfway between \(2\) and \(12\).
- Mean \(4\), halfway between \(-13\) and \(21\).
- Mean \(3\), the middle value and halfway between \(-9\) and \(15\).
- Mean \(47.5\), halfway between \(43\) and \(52\).
- Mean \(\frac{202}{3}=67\frac13\), between \(38\) and \(100\), slightly above \(64\); the deviations from this mean sum to zero.
- \(9\), found by \(\frac{5+13}{2}\).
- \(34\), found by \(\frac{26+42}{2}\).
- \(67\), found by \(\frac{57+77}{2}\).
- \(-28\), found by \(\frac{-34-22}{2}\).
- \(-21\), found by \(\frac{-45+3}{2}\).
- \(3\), found by \(\frac{-12+18}{2}\).
3. Set review — Find midpoints on coordinate graphs
When a graph gives you the endpoints, you can use the same intuitive method as in the main lesson: start at the first point and move halfway toward the second. Find the total horizontal and vertical changes, take half of each, and add those half-changes to your starting coordinates. This works for slanted, horizontal, and vertical segments.
Suppose one plotted segment runs from \(A=(-6,3)\) to \(B=(4,-5)\).
Step 1: Find the movement from A to B.
\[\Delta x=4-(-6)=10,\qquad \Delta y=-5-3=-8.\]Step 2: Go halfway along that movement, starting at A. Half of 10 is 5; half of \(-8\) is \(-4\).
\[M=\left(-6+\frac{10}{2},\;3+\frac{-8}{2}\right)=(-6+5,\;3-4)=\boxed{(-1,-1)}.\]You can check: moving from \(M\) to \(B\) also goes 5 units right and 4 units down.
The familiar midpoint formula is simply a shorter way to write this same reasoning:
\[\begin{aligned} x_M&=x_1+\frac{x_2-x_1}{2}=\frac{x_1+x_2}{2},\\ y_M&=y_1+\frac{y_2-y_1}{2}=\frac{y_1+y_2}{2}. \end{aligned}\]So you can either start and move halfway or average the endpoints—whichever is easier for the problem. On a graph, read each endpoint's coordinates carefully; counting grid squares can help check your result. If a diagram contains several segments, identify the two endpoints of each before calculating.
If \(A=(0,0)\), \(B=(8,0)\), and \(C=(8,6)\), and the drawn segments are \(AB\), \(BC\), and \(AC\), they have three different midpoints:
\[M_{AB}=(4,0),\qquad M_{BC}=(8,3),\qquad M_{AC}=(4,3).\]Each midpoint belongs to the specific endpoints of its own segment.
4. Set review — Locate points with parallel lines and similar triangles
From Sections 6.2–6.4, you know that a line parallel to one side of a triangle forms a smaller similar triangle. Corresponding sides are proportional, even when the triangle is not a right triangle.
In the diagram, \(DE=4\), \(DB=12\), and \(DC=9\). Because \(EF\) is parallel to \(BC\), the smaller triangle has one-third the corresponding side lengths of the larger:
\[\frac{DF}{DC}=\frac{DE}{DB}=\frac{4}{12},\qquad DF=9\cdot\frac13=\boxed{3}.\]So the parallel segment meets the vertical side at \(F=(0,3)\).
For coordinate problems, a useful rule follows: if a parallel line cuts a fraction \(t\) of one side measured from a triangle’s vertex, it cuts the same fraction of the other side from that same vertex. That is exactly the coordinate-partition idea from the main lesson.
In the figure, \(D=(-1,-1)\), \(B=(-1,8)\), \(C=(11,-1)\), and \(A=(-1,2)\). Draw \(AP\) parallel to \(BC\), with P on \(DC\).
Along the left side, \(DA=3\) and \(DB=9\), so the smaller triangle is at scale \(t=1/3\) from D. Therefore P is one-third of the way from D to C:
\[x_P=-1+\frac13(11-(-1))=3,\qquad y_P=-1+\frac13(-1-(-1))=-1.\]Thus \(\boxed{P=(3,-1)}\).
Let \(A=(2,1)\), \(B=(8,4)\), and \(C=(5,10)\). Choose \(P=(4,2)\) on \(AB\). The changes A to B and A to P are \((6,3)\) and \((2,1)\), so \(AP/AB=1/3\).
A line through P parallel to BC meets AC at Q, also one-third from A:
\[Q=(2,1)+\frac13(5-2,10-1)=\boxed{(3,4)}.\]Neither triangle is a right triangle. What matters is the parallelism and corresponding ratios.
Set — Check midpoints and proportional coordinates
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Use the midpoint worked examples for problems 13–18 and the parallel-line worked examples for problems 19–24.
Check your work — Set 13–24 answer key
- \((5,4)\).
- \((2,2)\).
- Vertical side: \((6,-\frac12)\); horizontal side: \((8,-4)\); diagonal: \((8,-\frac12)\).
- Horizontal side: \((\frac32,5)\); vertical side: \((4,\frac32)\); diagonal: \((\frac32,\frac32)\).
- \((6,9)\).
- \((\frac12,-\frac12)\).
- \((4,-2)\). Since \(\frac{DA}{DB}=\frac46=\frac23\), the intersection is \(\frac23\) of the way from \(D=(-2,-2)\) to \(C=(7,-2)\).
- \((\frac92,-1)\). Since \(E\) is halfway from \(C=(7,-1)\) to \(B=(7,9)\), the intersection is halfway along \(CD\), with \(D=(2,-1)\).
- \((-1,4)\). The fraction from \(D=(-4,4)\) toward \(B=(-4,1)\) is \(\frac13\), so take \(\frac13\) of \(DC\), where \(C=(5,4)\).
- \((5,2)\). With \(D=(-5,4),G=(1,4),C=(5,4)\), \(\frac{CG}{CD}=\frac4{10}\); take \(\frac25\) of the way from \(C\) to \(B=(5,-1)\).
- The two sides are divided in the same ratio: corresponding smaller triangles are similar by AA, so their corresponding sides are proportional; subtracting the smaller lengths from the full lengths gives the same ratio for the remaining parts.
- Yes, provided the coordinates and parallel-line condition are known. AA similarity and the proportional division of a segment do not require a right angle.
5. Go review — Proportions from parallel lines
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
When several parallel lines cut two transversals, the segments between the same pair of parallel lines have equal ratios. Keep their order consistent on both transversals. These exercises review the proportionality theorem from Section 6.4.
Two rays from the same vertex meet two parallel segments. On the upper ray, the distance from the vertex to the first parallel is 6 units and from the first to the second parallel is 4 units. On the lower ray, the corresponding distances are 9 units and \(x\). Find \(x\).
Match the near segments and the farther segments:
\[\frac{6}{4}=\frac{9}{x}\quad\Longrightarrow\quad 6x=36\quad\Longrightarrow\quad\boxed{x=6}.\]Check: both rays scale by \(9/6=3/2\), so \(4(3/2)=6\).
Three parallel lines cut two crossing transversals. The successive segments on the first transversal are 7 and 12. The corresponding successive segments on the other are \(x+1\) and \(2x+1\). Find \(x\) and both unknown segment lengths.
\[\begin{aligned} \frac{7}{12}&=\frac{x+1}{2x+1},\\ 7(2x+1)&=12(x+1),\\ 14x+7&=12x+12,\\ 2x&=5,\qquad x=\frac52. \end{aligned}\]The actual lengths are \(x+1=\frac72\) and \(2x+1=6\). Check: \((7/12)=(3.5/6)\).
Check your work — Go 25–26 answer key
- \(\frac8{12}=\frac5x\), so \(x=7.5\).
- Assuming the three cross-lines are parallel, \(\frac48=\frac3y\), so \(y=6\). The diagram does not mark parallelism.
Additional Learning Resources
These links support the skills of Section 6.6. More advanced material is identified so that it does not become an unintended prerequisite.
| Resource and link | What it teaches | Scope / What to skip |
|---|---|---|
| Khan Academy — Midpoint formula practice | Coordinate midpoints from given endpoints, including signs and noninteger answers. Particularly useful for Main 1–6 and Set 13–18. | Close match. Use the midpoint exercises directly. |
| Khan Academy — Dividing line segments: graphical | Short visual explanation of dividing a line segment in a ratio, matching Main 7–8. | Close match. Work with internal division of segments; external division is not needed. |
| Khan Academy — Divide line segments practice | Practice finding coordinates of a point at a given fraction of a segment. | Close match. Complements the graphical video. |
| OpenStax, College Algebra 2e — §2.1, “Using the Midpoint Formula” | Textbook derivation and worked coordinate examples. | Some extra material. Use only the subsection titled “Using the Midpoint Formula”; other parts of §2.1 discuss graphing, distance, and intercepts. |
Paul’s Online Math Notes and Professor Leonard were checked for a suitable direct match; no closely targeted lesson or lecture was verified for this day’s segment-partition topic, so neither is assigned here.
Teacher review — Substantive packet issues
1. Incorrect stated midpoint (Main task, Problem 2, p. 31)
The packet requests a midpoint of \((-3,6)\), despite using endpoints \((-9,6)\) and \((3,10)\) and describing \((-3,8)\) immediately above. This is a substantive mathematical typo because the stated midpoint is not halfway between the endpoints. Correction: change “(−3, 6)” to “(−3, 8).”
2. No other serious problem identified
The remaining tasks give enough information for their intended techniques. The ratio and parallel-line methods are mathematically well defined when corresponding segments are matched carefully.
Section 6.6 packet-to-lesson coverage
| Packet location | Skills covered | Where in this supplement |
|---|---|---|
| Main, pp. 30–32, 1–6 | Locate, prove, and generalize the midpoint | In class · Worked 1–2, congruent-triangle argument, midpoint formula |
| Main, pp. 32–33, 7–8 | Partition segments in 1:2 and 2:3 ratios | In class · Worked 3–5, partition rule |
| Ready, p. 34, 1–12 | Arithmetic mean; halfway values for positive and negative numbers | Section 2 · Worked 6–8 |
| Set, pp. 34–35, 13–18 | Midpoints from graphs and given endpoints; multiple segments | Section 3 · Worked 9–10 |
| Set, pp. 35–36, 19–24 | Use similar triangles and parallel sides to find points; generality beyond right triangles | Section 4 · Worked 11–13 |
| Go, p. 36, 25–26 | Proportions for parallel-line cuts | Section 5 · Go worked A–B |
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