MAT8 Q2W5D3: Triangle classification

By the end of the lesson, you will be able to:

• Compute straight-line distances between points on the coordinate plane in exact form and as rounded decimals with appropriate units.
• Classify a triangle from coordinates by comparing squared side lengths and identify the vertex of the right angle when it exists.
• Select the appropriate distance model (Euclidean or city-walk) for a context and communicate results clearly with units and rounding.

• Distance formula: Straight-line distance from differences in coordinates.
• Squared distance: Use of ${\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2$ before taking roots.
• Classification: right, acute, obtuse — determined by comparing squared side lengths.
• Largest squared side ($L$) and sum of the other two ($S$).
• Converse of the Pythagorean Theorem: Right if $L=S$.
• City-walk (Manhattan) distance: Axis-aligned path length $\left|\mathrm{\Delta x}\right|+\left|\mathrm{\Delta y}\right|$.

1. Compute the distance between $(0,0)$ and $(9,12)$. Give exact and decimal.
2. Simplify exactly: $\sqrt{72}$.
3. Explain why rounding square roots early can cause misclassification in a near-right triangle.

Show Answer

1) $d=\sqrt{9^2+{12}^2}=\sqrt{225}=15$. Decimal: $15.0$.

2) $\sqrt{72}=\sqrt{36·2}=6\sqrt{2}$.

3) In near-equality cases, rounding may flip equality to less-than or greater-than; comparing squared lengths avoids rounding and preserves exact order.

0) Essential question and objective cues

Essential Question: How can squared distances alone tell us whether a triangle is right, acute, or obtuse, without measuring angles?

Objective cues you will see:

• Objective 1 check - compute straight-line distances exactly and approximately.
• Objective 2 check - classify a triangle by comparing squared side lengths and name the right-angle vertex when it exists.
• Objective 3 check - select the correct distance model and communicate with units and rounding.

1) Launch - What makes a straight line useful in coordinates

You have already used the distance formula to measure the straight-line length between two points on the coordinate plane. Today you will go further. You will use distances not only to measure but also to reason about shapes, especially triangles. The big idea is this: if you know how long each side of a triangle is, you can tell whether it is right, acute, or obtuse by comparing squared lengths. You will also practice choosing exact forms and appropriate units, and you will decide whether Euclidean or city-walk distance fits a situation.

Bridge: You will classify by comparing squares, then apply this to real placements where a right angle matters for safety and design.

Guiding question 1.1 - Why might comparing squared side lengths help you decide if an angle is right instead of measuring angles directly?

Show Answer

Because the Pythagorean relationship in a right triangle is a statement about squared side lengths. If the largest squared side equals the sum of the other two squared sides, then the triangle is right by the converse of the Pythagorean Theorem. This avoids angle measurement entirely.

Mini-summary A
You can classify a triangle using only side-length comparisons. Squared lengths are central because the Pythagorean Theorem and its converse speak in terms of squares.
Objective 2 check - you can explain why comparing squared lengths is sufficient.

2) From distance to classification - the classification test

You will use a consistent three-step test.

Step T1. Compute all three squared side lengths from coordinates.

Given three points $A=(x_1,y_1)$, $B=(x_2,y_2)$, $C=(x_3,y_3)$, compute:

• ${\mathrm{AB}}^2={(x_2-x_1)}^2+{(y_2-y_1)}^2$
• ${\mathrm{BC}}^2={(x_3-x_2)}^2+{(y_3-y_2)}^2$
• ${\mathrm{CA}}^2={(x_1-x_3)}^2+{(y_1-y_3)}^2$

Step T2. Identify the largest squared side. Let the results be $p$, $q$, $r$. Let $L$ be the largest and $S$ the sum of the other two.

Step T3. Compare using a single inequality and name the vertex.

• If $L=S$, the triangle is right, and the right angle is at the vertex opposite the side whose squared length is $L$.
• If $L<S$, the triangle is acute.
• If $L>S$, the triangle is obtuse.

Error inoculation

• Do not compute ${(\mathrm{\Delta x}+\mathrm{\Delta y})}^2$. Use ${\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2$.
• Track signs carefully for each difference before squaring.
• Keep squares exact. Avoid early square roots.

Guiding question 2.1 - Why is it better to compare squared lengths instead of square roots when classifying?

Show Answer

Square roots can be messy to compute. Since the Pythagorean Theorem is about squares, comparing squared lengths is exact and avoids rounding. It also preserves the correct order because the square function is increasing on nonnegative numbers.

Guiding question 2.2 - If two vertices share the same x-coordinate, what special property does the segment between them have?

Show Answer

The segment is vertical. Its horizontal change is zero, so its squared length becomes the square of the vertical change only.

Mini-summary B
To classify a triangle from coordinates, compute three squared distances, take the largest one, then compare with the sum of the other two. No angles are needed.
Objective 2 check - you know the inequality test and how to name the right-angle vertex.

2.5) Contrast pair - sharpen the discrimination before practice

Case CP-1 (near-right acute): $A=(0,0)$, $B=(6,8)$, $C=(6,7)$. Think first for 30 seconds before opening.

Show Answer

Squared lengths: ${\mathrm{AB}}^2=100$, ${\mathrm{BC}}^2=1$, ${\mathrm{CA}}^2=85$. Largest is 100, others sum to 86, so $100>86$ and the triangle is obtuse.

Case CP-2 (right): $P=(1,1)$, $Q=(5,1)$, $R=(5,7)$. Think first for 30 seconds before opening.

Show Answer

Squared lengths: ${\mathrm{PQ}}^2=16$, ${\mathrm{QR}}^2=36$, ${\mathrm{RP}}^2=52$. Largest is 52, others sum to 52, so the triangle is right with right angle at $Q$.

Inductive micro-activity (notice - group - name)
List what stayed the same across the two cases, group the common features, then name the rule. Think first for 20 seconds before opening.

Show Answer

Notice: we always computed 3 squared side lengths and compared the largest to the sum of the other two. Group: “largest square” and “sum of other squares.” Name: If largest square equals the sum, the triangle is right; if less, acute; if greater, obtuse.

3) Skill builder - fast and accurate squared distances

You will practice computing squared distances efficiently while avoiding common arithmetic slips.

Technique S1. Organize differences in a tiny table

Pair	$\mathrm{\Delta x}$	$\mathrm{\Delta y}$	squared distance
AB	compute	compute	${\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2$
BC	compute	compute	same formula
CA	compute	compute	same formula

Guiding question 3.1 - Why is it safe to ignore absolute value bars for differences before squaring?

Show Answer

Because squaring removes sign. For any real number $k$, ${(-k)}^2=k^2$.

Practice S1

Classify triangle with $A=(0,0)$, $B=(6,3)$, $C=(2,9)$. Think first before opening the solution.

Show Answer

Compute squared lengths: ${\mathrm{AB}}^2=6^2+3^2=45$, ${\mathrm{BC}}^2={(2-6)}^2+{(9-3)}^2={(-4)}^2+6^2=52$, ${\mathrm{CA}}^2=2^2+9^2=85$. The largest is $85$. The sum of the other two is $45+52=97$. Since $85<97$, the triangle is acute.

Mini-summary C
Organize differences carefully. Compare squared lengths to classify: equal means right, less means acute, greater means obtuse.
Objective 2 check.

4) Concept focus - why the classification test works

Think about a triangle where sides opposite the angles are labeled $a$, $b$, $c$, and suppose $c$ is the longest side. For a right triangle with right angle opposite $c$, the Pythagorean Theorem states $c^2=a^2+b^2$.

For a non-right triangle, geometry shows that if the angle opposite $c$ is acute, then $c^2<a^2+b^2$; if it is obtuse, then $c^2>a^2+b^2$. This is why the squared test gives an exact classification without measuring any angle.

Guiding question 4.1 - If you know three coordinates, why is it logical to treat the largest squared distance as the candidate for the hypotenuse in the right-triangle test?

Show Answer

Because the hypotenuse is the longest side in a right triangle. If a right angle exists, it must face the longest side. Therefore the Pythagorean equality can only be checked with the largest squared distance on one side of the equality.

Mini-summary D
Pythagorean comparisons extend beyond right triangles. The largest side controls the classification.
Objective 2 check.

5) Checkpoint 1 - independent classification

Classify the triangle with $P=(2,1)$, $Q=(8,-2)$, $R=(-1,7)$. Work in your notebook using the T1–T3 steps. Think first before opening.

Show Answer

Compute squared lengths. ${\mathrm{PQ}}^2=6^2+{(-3)}^2=45$, ${\mathrm{QR}}^2={(-9)}^2+9^2=162$, ${\mathrm{RP}}^2=3^2+{(-6)}^2=45$. Largest is $162$. Sum of the other two is $90$. Since $162>90$, the triangle is obtuse.

Mini-summary E
You compared squared lengths and decided obtuse because the largest squared side exceeded the sum of the other two.
Objective 2 check.

6) Real-world application 1 - sports field safety triangle

Bridge: You have the square-comparison test. Use it to certify safety where a right angle matters.

A coach places three pylons at $(5,2)$, $(11,2)$, and $(11,9)$ meters on a training grid. The angle at the first pylon must be a right angle. Confirm the triangle is right, and identify where the right angle is.

Work it out
Use the classification test. Think first before opening.

Show Answer

Compute squared distances. Label points $A=(5,2)$, $B=(11,2)$, $C=(11,9)$. ${\mathrm{AB}}^2=6^2+0=36$, ${\mathrm{BC}}^2=0+7^2=49$, ${\mathrm{CA}}^2={(-6)}^2+{(-7)}^2=36+49=85$. Largest squared distance is $85$. The sum of the other two is $36+49=85$. Since they are equal, the triangle is right with the right angle at point $B$.

Units checkpoint
Write the unit used above and explain why squared distances are unit-consistent before square roots.

Show Answer

Meters. Each squared distance uses meters squared, but the equality and comparisons are unit-consistent. If a final length is needed, take square roots and report meters.

Mini-summary F
Right triangles are detected when the largest squared side equals the sum of the other two. You confirmed the safety requirement and accounted for units.
Objective 1 check and Objective 2 check.

7) Real-world application 2 - drone navigation and city vs straight-line

Bridge: Now that you can classify by squares, choose the correct distance model when the path is constrained.

A drone must travel from a base at $(0,0)$ to a waypoint at $(120,90)$ meters. The straight-line distance is one measure. If winds and regulations force the drone to travel in a two-leg path, first the full horizontal, then the full vertical, that is like city-walk distance.

Task
Compute both distances. Then calculate how much longer the city-walk path is than the straight line. Think first before opening.

Show Answer

Straight-line distance: $d=\sqrt{{120}^2+{90}^2}=\sqrt{22500}=150$ $\text{meters}$. City-walk path: $120+90=210$ meters. Extra distance is $210-150=60$ meters.

Guiding question 7.1 - Which distance should a battery usage estimate use, and why?

Show Answer

Battery usage depends on the actual path. If the drone must fly the two-leg route, the estimate should use 210 meters. If it can fly straight, then 150 meters is appropriate. Use the constraint that truly applies.

Model-choice link to classification
Which model would overestimate a hypotenuse and why does that matter for safety margins?

Show Answer

City-walk distance is longer than Euclidean. If a safety margin requires the shortest path (hypotenuse) to be below a limit, city-walk would overestimate. That can protect safety margins but may over-restrict design if you need the exact diagonal.

Mini-summary G
Use Euclidean distance for unconstrained straight paths. Use city-walk for constrained axis-aligned movement. Choose the model that matches reality.
Objective 3 check.

8) Real-world application 3 - scale drawing decisions

On a campus map, grid squares are 1 cm by 1 cm, with a scale of $1\text{cm}=50\text{m}$. Two buildings on the sketch sit at $(1,4)$ and $(6,10)$. Find the real straight-line distance. Give an exact expression first, then a rounded decimal with proper units. Think first before opening.

Show Answer

Map distance in centimeters: $d=\sqrt{{(6-1)}^2+{(10-4)}^2}=\sqrt{5^2+6^2}=\sqrt{61}$ $\text{cm}$. Real distance: multiply by 50 m per cm. $50\times \sqrt{61}\approx 390.5$ $\text{m ≈ 391 m}$.

Units checkpoint
Explain why it is good practice to keep the radical exact until after scaling.

Show Answer

Rounding early can carry errors into the scaled result. Keeping $\sqrt{61}$ exact before multiplying preserves precision.

Mini-summary H
When maps have scales, compute the geometric distance on the map first, then convert to real units. Postpone rounding until the final number.
Objective 1 check and Objective 3 check.

9) Deep practice - classification across cases

You will now classify several triangles. Record all work.

Case D1

$A=(2,3)$, $B=(-4,3)$, $C=(-4,-5)$ Think first before opening.

Show Answer

Compute squared distances. ${\mathrm{AB}}^2={(-6)}^2+0=36$, ${\mathrm{BC}}^2=0+{(-8)}^2=64$, ${\mathrm{CA}}^2=6^2+8^2=100$. Largest is $100$. Sum of the other two is $100$. It is a right triangle. The right angle is at $B$.

Case D2

$P=(1,2)$, $Q=(5,7)$, $R=(9,1)$ Think first before opening.

Show Answer

${\mathrm{PQ}}^2=4^2+5^2=41$, ${\mathrm{QR}}^2=4^2+{(-6)}^2=52$, ${\mathrm{RP}}^2={(-8)}^2+1^2=65$. Largest is $65$. Sum of the other two is $93$. Since $65<93$, it is acute.

Case D3

$U=(4,4)$, $V=(10,1)$, $W=(1,13)$ Think first before opening.

Show Answer

${\mathrm{UV}}^2=6^2+{(-3)}^2=45$, ${\mathrm{VW}}^2={(-9)}^2+{12}^2=225$, ${\mathrm{WU}}^2=3^2+{(-9)}^2=90$. Largest is $225$. Sum of the other two is $135$. Since $225>135$, the triangle is obtuse.

Mini-summary I
The test is reliable across many coordinate positions. Always select the largest squared side for the comparison.
Objective 2 check.

10) Error spotting - common pitfalls and quick checks

1. Adding differences first, then squaring the total. The formula requires squaring each difference separately then adding.
2. Dropping a negative sign when subtracting y-coordinates.
3. Forgetting that units matter when converting from a scaled map.
4. Rounding too early, then using the rounded number in further steps.

Guiding question 10.1 - If your classification depends on small differences, how can rounding errors mislead you?

Show Answer

Rounding square roots early can change which side appears largest or change whether equality seems to hold. Comparing squared lengths avoids rounding and preserves exact comparisons.

Mini-summary J
Prevent mistakes by keeping squared comparisons exact, tracking signs carefully, and delaying rounding.
Objective 1 check and Objective 2 check.

11) Design problem - diagonal constraint and clearance

A rectangular storage area is being designed in a lab. The floor plan rectangle must fit a diagonal conduit that can be at most $20\text{m}$ long. If the rectangle is proposed to be $12\text{m}$ by $16\text{m}$, check the diagonal and state whether it meets the requirement. Think first before opening.

Show Answer

$d=\sqrt{{12}^2+{16}^2}=\sqrt{400}=20$ $\text{m}$, so it meets the at most 20 m constraint.

Mini-summary K
Rectangular diagonals are direct applications of the distance formula. Constraints can be checked quickly and exactly.
Objective 1 check.

12) Mixed application - plan a route with a safety right angle

An events team places three stakes for a temporary walkway: $S=(1,1)$, $T=(7,1)$, $U=(7,8)$. They need a right turn at T. Confirm and justify. Think first before opening.

Show Answer

${\mathrm{ST}}^2=6^2+0=36$, ${\mathrm{TU}}^2=0+7^2=49$, ${\mathrm{US}}^2={(-6)}^2+{(-7)}^2=36+49=85$. Largest squared side is $85$. The other two sum to $85$. Right angle at $T$.

Mini-summary L
You validated a safety turn using the classification procedure. Squared distances are enough.
Objective 2 check.

13) Strategy - when to use exact radical and when to round

• Use exact radicals when you will classify or compare.
• Use decimals with units when you must communicate a real measurement, often with a specified accuracy such as nearest tenth or meter.

Guiding question 13.1 - If you compute a diagonal as $\sqrt{128}$, what exact simplification helps before rounding?

Show Answer

$\sqrt{128}=\sqrt{64·2}=8\sqrt{2}$.

Mini-summary M
Simplify radicals exactly before rounding. This keeps results clear and reduces calculator mistakes.
Objective 1 check.

14) Independent practice checkpoint - mixed tasks

Work these without a calculator first. Then check.

Task P1 - classify
$J=(-2,6)$, $K=(5,6)$, $L=(5,-2)$ Think first before opening.

Show Answer

${\mathrm{JK}}^2=7^2+0=49$, ${\mathrm{KL}}^2=0+{(-8)}^2=64$, ${\mathrm{LJ}}^2={(-7)}^2+8^2=113$. Largest is $113$. The other two sum to $113$. Right triangle, right angle at $K$.

Task P2 - distance and units
On a diagram with 1 cm representing 0.5 km, points are $(3,1)$ and $(10,5)$. Find the real distance. Think first before opening.

Show Answer

Map distance: $\sqrt{7^2+4^2}=\sqrt{65}$ cm. Real distance: $\frac{1}{2}\sqrt{65}$ km $\approx 4.03$ km.

Task P3 - city vs straight
From $(1,2)$ to $(10,14)$, compare Euclidean and city-walk distances. Think first before opening.

Show Answer

Straight-line: $\sqrt{9^2+{12}^2}=15$. City-walk: $9+12=21$.

Mini-summary N
You practiced classification, unit conversion, and model choice. Keep exact comparisons as squares when classifying.
Objective 1 check, Objective 2 check, Objective 3 check.

15) Reflection pause - think like a designer

Suppose you must guarantee a right angle at one vertex of a triangle in a plan. Explain in your notebook how you would place three points with integer coordinates to achieve a right angle without trial and error. Think first before opening.

Show Answer

Place one side horizontal and one side vertical. For example, choose $(2,5)$, $(9,5)$, $(9,11)$. The angle at $(9,5)$ is right because horizontal and vertical segments meet.

Mini-summary O
Placing points to create a right angle is easy when you align one pair horizontally and the other vertically. The classification test will confirm it.
Objective 2 check.

16) Extension - triangle class from a real data snippet

A survey map lists three landmarks as approximate coordinates in meters:

• Water tower $(-30,45)$
• Gate $(10,45)$
• Trailhead $(10,5)$

Classify the triangle and compute the exact area to check signage placement needs. Hint for area: right triangles have area $\frac{1}{2}\times \mathrm{leg}\times \mathrm{leg}$. Think first before opening.

Show Answer

Squared distances: ${\mathrm{WT}-G}^2={40}^2+0=1600$, ${G-T}^2=0+{(-40)}^2=1600$, ${T-\mathrm{WT}}^2={(-40)}^2+{40}^2=3200$. Largest is $3200$. Sum of other two is $3200$. Right triangle. Legs are 40 m by 40 m, so area is $\frac{1}{2}\times 40\times 40=800$ square meters.

Mini-summary P
With coordinates you can classify triangles and compute areas without drawing angles or using special tools.
Objective 2 check and Objective 3 check.

17) Think aloud - create your own classification test kit

Build a repeatable checklist for any triangle from coordinates.

Proposed kit

1. Write all three points clearly.
2. Make a difference table for each pair.
3. Compute three squared distances carefully.
4. Circle the largest squared distance.
5. Add the other two squared distances.
6. Compare largest with the sum.
7. State the class and identify the vertex of the right angle if applicable.
8. If required, compute a real distance for communication with units and rounding.

Guiding question 17.1 - Why is a checklist valuable in testing situations?

Show Answer

It reduces errors under time pressure, ensures consistent steps, and makes it easier to catch missing calculations or sign mistakes.

Mini-summary Q
A short checklist stabilizes your method. Consistent steps lead to consistent accuracy.
Objective 2 check.

18) Robust understanding - justify the right triangle test

Sketch any triangle and label the longest side length $c$. Build a geometric argument that if the angle opposite $c$ were right, then $c^2=a^2+b^2$, and contrast with the acute and obtuse cases. Think first before opening.

Show Answer

For a right triangle, the Pythagorean Theorem holds exactly. If the included angle is acute, projecting side lengths shows the opposite side is too short to satisfy equality, so $c^2<a^2+b^2$. If the angle is obtuse, the opposite side must be longer than what equality would give, so $c^2>a^2+b^2$.

Mini-summary R
The squared comparison captures how angles open or close a triangle. Equality marks a right angle.
Objective 2 check.

19) Final checkpoint - multi-step real problem

A rectangular plaza will use a diagonal light cable across two corners. The plaza scale drawing uses 1 cm to 2 m. Corners on the plan are at $(2,1)$ and $(15,11)$ for opposite corners. a) Find the plan diagonal length in centimeters. b) Convert to real meters. c) If the cable is sold in whole meters only, what length should be purchased? d) If a safety rule requires the shortest approach to a utility box at $(15,1)$, decide if the triangle formed by the two corners and the box is right, acute, or obtuse. Think first before opening.

Show Answer

a) Plan diagonal: $d=\sqrt{{(15-2)}^2+{(11-1)}^2}=\sqrt{{13}^2+{10}^2}=\sqrt{269}$ cm.

b) Real meters: multiply by 2. $2\times \sqrt{269}\approx 32.802$ m.

c) Purchase 33 m to cover the length.

d) Classification with points $C=(2,1)$, $D=(15,11)$, $B=(15,1)$. ${\mathrm{CD}}^2=269$, ${\mathrm{DB}}^2=100$, ${\mathrm{BC}}^2=169$. Largest is $269$. Sum of the other two is $269$. It is a right triangle with the right angle at $B$.

Mini-summary S
You combined distance, unit conversion, and classification in a single problem. You justified decisions with exact comparisons.
Objective 1 check, Objective 2 check, Objective 3 check.

20) Quick self-check - concept statements

Decide if each statement is true or false. Then explain briefly.

1. If the largest squared side is less than the sum of the other two squared sides, the triangle is acute.
2. City-walk distance is always at least as large as Euclidean straight-line distance between the same points.
3. When classifying triangles from coordinates, you must first compute exact square roots.
4. For a rectangle with side lengths $w$ and $h$, the diagonal is $\sqrt{w^2+h^2}$.

Show Answer

1) True. 2) True. 3) False, you can compare squared distances directly. 4) True, by the distance formula or Pythagoras.

Mini-summary T
Key relations: compare squares to classify, and match the distance model to the scenario.
Objective 2 check and Objective 3 check.

21) Vocabulary retrieval prompts

Answer these in your notebook, then open to verify.

• What does distance formula mean in your own words?
  Show Answer

  A rule that gives the straight-line length between two coordinate points by squaring and adding the horizontal and vertical differences, then taking a square root if a length is needed.

• What is the converse of the Pythagorean Theorem?
  Show Answer

  If the largest squared side equals the sum of the other two squared sides, the triangle is a right triangle.

• What is city-walk (Manhattan) distance?
  Show Answer

  A path length restricted to axis-aligned segments, found by adding the absolute horizontal change and the absolute vertical change.

22) Metacognitive strategy selector

When should you compute roots, compare squares, or use structure directly?

Show Answer

Compare squares for classification. Compute roots when you must communicate a length with units. Use structure when segments are clearly horizontal or vertical so you can reason quickly without full computation.

23) Wrap-up - what you can now do

• Compute straight-line distances exactly and approximately.
• Classify triangles using squared side comparisons and name the right-angle vertex when applicable.
• Choose the correct model of distance for real situations and communicate results with appropriate units and rounding.

References

• Euclid. Elements, Book I, Proposition 47 and its converse.
• Maor, E. The Pythagorean Theorem: A 4,000-Year History.
• OpenStax. Geometry and Algebra sections on distance in the coordinate plane.
• Posamentier, A., and Lehmann, I. The Secrets of Triangles.

Example 1 — Distance, exact and decimal

Problem. Find the straight-line distance between $(-3,4)$ and $(5,-2)$. Give the exact value and a decimal to the nearest tenth.

Show Answer

1. Compute differences: $\mathrm{\Delta x}=5--3=8$, $\mathrm{\Delta y}=-2-4=-6$.
2. Distance: $d=\sqrt{8^2+{-6}^2}=\sqrt{100}=10$.
3. Exact value: $10$. Decimal: $10.0$.

Example 2 — Classify a triangle and name the right-angle vertex

Problem. Classify the triangle with $A=(1,2)$, $B=(7,2)$, $C=(7,8)$. If right, name the vertex of the right angle.

Show Answer

1. Squared lengths:
   ${\mathrm{AB}}^2={(7-1)}^2+{(2-2)}^2=6^2+0=36$
   ${\mathrm{BC}}^2=0+6^2=36$
   ${\mathrm{CA}}^2={(1-7)}^2+{(2-8)}^2={-6}^2+{-6}^2=72$
2. Largest is $72$. Others sum to $72$.
3. Since equal, the triangle is right. The right angle is at the vertex opposite side $\mathrm{CA}$, which is point $B$.

Example 3 — Classify acute vs obtuse

Problem. Classify the triangle with $A=(0,0)$, $B=(4,5)$, $C=(7,1)$.

Show Answer

1. ${\mathrm{AB}}^2=41$
2. ${\mathrm{BC}}^2=25$
3. ${\mathrm{CA}}^2=50$
4. Largest $50$; others sum $66$. Since $50<66$, the triangle is acute.

Example 4 — Scale drawing to real distance

Problem. On a map with scale $1\text{cm}=200\text{m}$, two points are $(2,3)$ and $(9,11)$. Find the real straight-line distance. Give the exact form first, then a rounded value to the nearest meter.

Show Answer

1. Map distance (cm): $d=\sqrt{{(9-2)}^2+{(11-3)}^2}=\sqrt{7^2+8^2}=\sqrt{113}$ cm.
2. Real distance (m): $200\times \sqrt{113}$. Approximate: $\sqrt{113}\approx 10.630$, so $200\times 10.630=2126$ m.
3. Answer: exact $200\sqrt{113}$ m; rounded $2126$ m.

Example 5 — City-walk vs straight-line

Problem. From $(-8,6)$ to $(5,-2)$, find the Euclidean distance, the city-walk distance, and how much longer the city-walk path is.

Show Answer

1. Differences: $\mathrm{\Delta x}=13$, $\mathrm{\Delta y}=-8$.
2. Euclidean: $\sqrt{233}\approx 15.264$.
3. City-walk: $21$.
4. Extra length: $21-\sqrt{233}\approx 5.736$.

1) Distance — exact and decimal

Find the distance between $(-7,3)$ and $(5,-9)$. Give the exact value and a decimal to the nearest tenth.

Show Answer

Differences: $\mathrm{\Delta x}=12$, $\mathrm{\Delta y}=-12$. Distance: $d=\sqrt{288}=12\sqrt{2}$ ≈ $17.0$.

2) Classify and name the right-angle vertex

Classify the triangle with $A=(2,-1)$, $B=(8,5)$, $C=(2,5)$. If right, name the vertex of the right angle.

Show Answer

${\mathrm{AB}}^2=72$, ${\mathrm{BC}}^2=36$, ${\mathrm{CA}}^2=36$. Largest is $72$; sum of others $72$. Right triangle. Right angle at the vertex opposite side $\mathrm{AB}$, which is $C$.

3) Classify obtuse or acute

Classify the triangle with $P=(-3,4)$, $Q=(5,7)$, $R=(9,-2)$.

Show Answer

${\mathrm{PQ}}^2=73$, ${\mathrm{QR}}^2=97$, ${\mathrm{RP}}^2=180$. Largest $180$; others sum $170$. Since $180>170$, obtuse.

4) Scale drawing to real units

On a map with scale $1\text{cm}=0.2\text{km}$, points are $(1,2)$ and $(10,7)$. Find the real distance. Give exact and a decimal to the nearest hundredth.

Show Answer

Map distance: $\sqrt{9^2+5^2}=\sqrt{106}$ cm. Real distance: $0.2\times \sqrt{106}\approx 2.06$ km.

5) City-walk vs straight-line

Between $(3,-4)$ and $(-9,10)$, find the Euclidean distance, the city-walk distance, and the extra length of the city path.

Show Answer

Differences: $\mathrm{\Delta x}=-12$, $\mathrm{\Delta y}=14$. Euclidean: $\sqrt{340}=2\sqrt{85}\approx 18.44$. City-walk: $26$. Extra: $26-2\sqrt{85}\approx 7.56$.

6) Rectangle diagonal check

A storage room must have a diagonal at most $17\text{m}$. Proposed size is $15\text{m}$ by $8\text{m}$. Does it meet the requirement?

Show Answer

Diagonal: $\sqrt{289}=17$ $\text{m}$. Meets the at most 17 m limit exactly.

7) Right-angle verification and vertex

For $J=(-4,-1)$, $K=(2,-1)$, $L=(2,6)$, classify the triangle and name the right-angle vertex if right.

Show Answer

${\mathrm{JK}}^2=36$, ${\mathrm{KL}}^2=49$, ${\mathrm{LJ}}^2=85$. Since $85=36+49$, right triangle. Right angle at the vertex opposite side $\mathrm{LJ}$, which is $K$.

8) Classify acute

Classify the triangle with $A=(0,0)$, $B=(6,2)$, $C=(3,7)$.

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${\mathrm{AB}}^2=40$, ${\mathrm{BC}}^2=34$, ${\mathrm{CA}}^2=58$. Largest $58$, others sum $74$. Since $58<74$, acute.

9) Construct a right triangle then compute the hypotenuse

Guarantee a right angle at $(4,-2)$ by placing integer-coordinate points so one leg is length 5 and the other is length 12. Give the third vertex coordinates, the hypotenuse length, and the classification.

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Choose horizontal leg to $(9,-2)$ (length 5) and vertical leg to $(4,10)$ (length 12). The third side between these two has $\mathrm{\Delta x}=-5$, $\mathrm{\Delta y}=12$, so hypotenuse $\sqrt{25+144}=\sqrt{169}=13$. Right triangle with right angle at $(4,-2)$.

10) Confirm right triangle and find the hypotenuse length

For $A=(0,0)$, $B=(12,9)$, $C=(0,9)$, classify the triangle and state the hypotenuse length.

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${\mathrm{AB}}^2=225$, ${\mathrm{BC}}^2=144$, ${\mathrm{CA}}^2=81$. Since $225=144+81$, right triangle. Hypotenuse is $\mathrm{AB}$ with length $\sqrt{225}=15$.

1) True/False

Comparing squared side lengths (not square roots) is sufficient to classify a triangle from coordinates.

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True.

2) True/False

If the largest squared side $L$ equals the sum of the other two squared sides $S$, the triangle is right and the right angle is at the vertex opposite the side with squared length $L$.

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True.

3) Classify and name the right-angle vertex

$A=(-2,5)$, $B=(6,5)$, $C=(6,11)$.

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${\mathrm{AB}}^2=64$, ${\mathrm{BC}}^2=36$, ${\mathrm{CA}}^2=100$. Since $64+36=100$, right at $B$.

4) Distance (exact and decimal to nearest tenth)

Between $(-4,-3)$ and $(9,5)$.

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$\mathrm{\Delta x}=13$, $\mathrm{\Delta y}=8$; $d=\sqrt{233}\approx 15.3$.

5) Scale drawing to real distance

Scale: $1\text{cm}=\frac{1}{4}\text{km}$. Points $(2,8)$, $(11,2)$. Real distance.

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Map: $\sqrt{117}$ cm; Real: $\frac{1}{4}\sqrt{117}$ km $\approx 2.70$ km.

6) City-walk vs Euclidean

From $(3,-4)$ to $(-5,7)$: Euclidean distance, city-walk distance, and extra length.

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$\mathrm{\Delta x}=-8$, $\mathrm{\Delta y}=11$; Euclidean $\sqrt{185}\approx 13.60$; City-walk $19$; Extra $19-\sqrt{185}\approx 5.40$.

7) Classify from squared lengths only

A triangle has squared side lengths $85$, $52$, $45$. Classify.

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Largest $85$; others sum $97$. Since $85<97$, acute.

8) Construct a right triangle (design)

Guarantee a right angle at $(7,2)$ using integer coordinates for the third point with the segment to $(1,2)$ as a horizontal leg.

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Example: choose $(7,9)$. Then one leg is horizontal and one is vertical, so the angle at $(7,2)$ is right.

9) Rounding and units

A computed straight-line distance is $12.47$ meters. Round to the nearest meter.

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$12$ meters.

10) Correct a formula error

A student used ${(\mathrm{\Delta x}+\mathrm{\Delta y})}^2$ for points $(-1,4)$, $(6,-2)$. Give the correct distance.

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Correct: $d=\sqrt{7^2+{-6}^2}=\sqrt{85}$.

11) Verify right triangle and vertex

$E=(0,0)$, $F=(8,6)$, $G=(0,6)$.

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${\mathrm{EF}}^2=100$, ${\mathrm{FG}}^2=64$, ${\mathrm{GE}}^2=36$. Since $64+36=100$, right at $F$.

12) Which pair is farther (Euclidean)

Compare distances: A: $(-3,-3)$ to $(9,9)$; B: $(-10,4)$ to $(4,-8)$.

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A: $d^2=288$; B: $d^2=340$. Pair B is farther.

13) Model choice (scenario)

A truck must use grid streets from $(2,-1)$ to $(-9,7)$. Which model applies and what path length should be used for fuel planning?

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City-walk model; distance $|-11|+\left|8\right|=19$ units.

14) Simplify exactly

Simplify $\sqrt{180}$.

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$\sqrt{180}=\sqrt{36·5}=6\sqrt{5}$.

15) Classify (aim for obtuse)

$A=(0,0)$, $B=(6,8)$, $C=(5,7)$.

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${\mathrm{AB}}^2=100$, ${\mathrm{BC}}^2=2$, ${\mathrm{CA}}^2=74$. Largest $100$; others sum $76$. Since $100>76$, obtuse.

🔒 Answer Key (condensed)

Show Answer

1. True
2. True
3. Right; vertex B
4. $\sqrt{233}$ ≈ 15.3
5. $\frac{1}{4}\sqrt{117}$ km ≈ 2.70 km
6. Euclid ≈ 13.60; City = 19; Extra ≈ 5.40
7. Acute
8. Example: $(7,9)$
9. 12 m
10. $\sqrt{85}$
11. Right; vertex F
12. Pair B
13. City-walk; 19 units
14. $6\sqrt{5}$
15. Obtuse

1) Build-your-own right triangle (integer grid)

Create a right triangle with integer coordinates whose legs are different lengths and whose hypotenuse length is exactly $13$ units. Give the three coordinates and verify by squared-length comparison.

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One choice (a 5-12-13 triple centered anywhere): Let $A=(2,1)$, $B=(7,1)$ (horizontal leg 5), and $C=(7,13)$ (vertical leg 12). Squared lengths: ${\mathrm{AB}}^2=25$, ${\mathrm{BC}}^2=144$, ${\mathrm{CA}}^2=169$. Since $25+144=169$, right triangle. Hypotenuse length $\sqrt{169}=13$.

2) Invariance check (translation and rotation)

Explain why the triangle classification by comparing squared side lengths is unchanged if you translate all three points by the same vector or rotate the entire figure about the origin.

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Translation: Squared distances use differences $(x_2-x_1)$ and $(y_2-y_1)$. Adding the same constants to both coordinates cancels in differences, so each squared distance is unchanged; therefore the comparison is unchanged. Rotation by angle $\theta$: Rotations preserve Euclidean distances. Since every squared distance is preserved, the largest-vs-sum comparison is preserved. Thus the classification remains the same.

3) Minimal change to flip type (right → acute/obtuse)

Start with a right triangle $A=(0,0)$, $B=(9,0)$, $C=(9,12)$. Move only point $C$ by one grid unit to make the triangle (a) acute and (b) obtuse. Verify using squared lengths.

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Original squared lengths: ${\mathrm{AB}}^2=81$, ${\mathrm{BC}}^2=144$, ${\mathrm{CA}}^2=225$. Right since $81+144=225$.

(a) For acute, move off the vertical leg: e.g., to $(8,9)$. Then ${\mathrm{BC}}^2=1^2+3^2=10$, ${\mathrm{CA}}^2=8^2+9^2=145$; sum smaller $81+10=91$, and $145<91$ is false, so adjust to $(8,9)$ gives largest $145$ and others sum $91$, hence not acute; to achieve acute in one step, choose a point that reduces the largest side enough: $(8,10)$ makes ${\mathrm{BC}}^2=5$, ${\mathrm{CA}}^2=164$, sum smaller $86$; still not acute. A reliable acute move is to place C so that all three squared sides are balanced: e.g., $(8,8)$ yields ${\mathrm{BC}}^2=\mathrm{(1^2+8^2)\; =\; 65}$, ${\mathrm{CA}}^2=\mathrm{(8^2+8^2)\; =\; 128}$, sum smaller $81+65=146$, and $128<146$ → acute.

(b) For obtuse, move slightly off the leg so the largest squared side exceeds the sum of the other two: e.g., $(8,11)$ gives ${\mathrm{BC}}^2=2$, ${\mathrm{CA}}^2=185$, sum smaller $83$, and $185>83$ → obtuse.

4) Choose the correct model (design constraint)

A robot must move from loading dock $(2,3)$ to station $(23,18)$. If aisles force horizontal/vertical motion only, which distance model should be used for timing and how much longer is this path than the straight-line diagonal?

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City-walk model for timing: $|23-2|+|18-3|=36$. Straight-line: $\sqrt{{21}^2+{15}^2}=\sqrt{666}$ $\approx 25.806$. Extra length: $36-\sqrt{666}\approx 10.194$ units.

5) Design to spec (bounded box)

Within the square $0\le x\le 10$, $0\le y\le 10$, place three integer-coordinate points so that:
- Triangle 1 is right.
- Triangle 2 is acute.
- Triangle 3 is obtuse.
For each, list coordinates and show the squared-length comparison.

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Right: $R=(2,2)$, $S=(8,2)$, $T=(8,7)$. Squared: ${\mathrm{RS}}^2=36$, ${\mathrm{ST}}^2=25$, ${\mathrm{TR}}^2=61$. Since $36+25=61$, right at $S$.

Acute (final set): $A=(1,1)$, $B=(4,5)$, $C=(6,3)$; then ${\mathrm{AB}}^2=25$, ${\mathrm{BC}}^2=8$, ${\mathrm{CA}}^2=29$; largest $29$, sum smaller $33$, and $29<33$ → acute.

Obtuse (final set): $O=(1,1)$, $P=(9,2)$, $Q=(3,3)$. Then ${\mathrm{OP}}^2=65$, ${\mathrm{PQ}}^2=37$, ${\mathrm{QO}}^2=8$; largest $65$, others sum $45$, so obtuse.

A) 3–2–1 Reflection

1. Three things I understand confidently now about classifying triangles from coordinates.
2. Two places I still make mistakes or feel unsure.
3. One question I want to ask my future self, plus a plan to answer it.

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Sample responses (guide only):

1. Three understands: (i) I can compute three squared side lengths from coordinates using ${\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2$. (ii) I can classify by comparing the largest squared side $L$ with the sum of the other two $S$: right if $L=S$, acute if $L<S$, obtuse if $L>S$. (iii) I can choose Euclidean vs. city-walk distance depending on whether motion is unconstrained or aisle/grid-restricted.
2. Two confusions: (i) I sometimes drop a sign when computing $\mathrm{\Delta y}$. (ii) I sometimes round too early instead of comparing squares.
3. One question + plan: “How do I quickly spot the right-angle vertex?” Plan: After I find the largest squared side, I will mark the vertex opposite that side as the right-angle candidate and verify with $c^2=a^2+b^2$.

B) Short Metacognitive Write-up (6–8 sentences)

Prompt: Explain how comparing squared lengths, instead of square roots, helps you avoid mistakes. Include one example where rounding a square root too early could change your classification.

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Sample points to include: Comparing squares keeps all values exact and avoids rounding noise; since the square function is increasing on nonnegative numbers, order is preserved. In a near-right case where $L$ is very close to $S$, rounding the square roots could flip equality to a “less than” or “greater than,” changing a right triangle into acute or obtuse incorrectly. Therefore I compare $a^2$, $b^2$, and $c^2$ directly, and take square roots only when I must report a final length with units.

C) Quick Checklist — My Process (tick in your notebook)

• ☐ I set up a small differences table for all three sides.
• ☐ I computed three squared distances with careful signs.
• ☐ I identified the largest squared distance and labeled the opposite vertex.
• ☐ I compared largest vs. sum ($L$ vs. $S$) without taking square roots.
• ☐ I stated the triangle type and, if right, the right-angle vertex.
• ☐ I chose Euclidean vs. city-walk to match the scenario.
• ☐ I used exact forms, then rounded with units only at the end if needed.

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Tip: Any unchecked box becomes your next practice target. For accuracy, prioritize “careful signs” and “compare squares.”

D) Snapshot Rubric — Where am I now

Level	Description
4	I consistently compute all three squared distances correctly, classify accurately using $L$ vs. $S$, and justify the right-angle vertex. I select the correct distance model with units.
3	I usually compute and classify correctly; I sometimes slip on signs or units but can self-correct using the checklist.
2	I can compute some squared distances, but I mix up which side is largest or I take square roots too early.
1	I need support setting up differences and deciding the model (Euclidean vs. city-walk).

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Self-evidence examples: “In Problem P1 I kept squared forms and got an acute classification since $L<S$.” “In the drone task I used city-walk for the constrained path and reported units.”

E) One-minute Plan — My next step

Write one concrete step you will do next time (e.g., “Always box the largest squared side before comparing,” or “Write units at the end only.”).

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Examples: “I will compute and list ${\mathrm{AB}}^2$, ${\mathrm{BC}}^2$, ${\mathrm{CA}}^2$ in the same order each time and circle the largest.” “I will keep radicals exact until the final communication step.”