MAT8 Q2W5D4: Classifying Triangles in the Coordinate Plane Using the Distance Formula

🎯 Learning Goals

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

1. Solve at least 4 real-world problems on the coordinate plane that require computing straight-line distances using $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$, reporting exact values and rounded results with correct units and specified precision (e.g., nearest tenth).
2. Classify triangles from three coordinates by comparing squared side lengths—right if $L=S$, acute if $L<S$, obtuse if $L>S$—and justify the decision in writing using computed values of $a^2$, $b^2$, $c^2$.
3. Choose the appropriate distance model for a context—Euclidean vs. city-walk $\left|\mathrm{\Delta x}\right|+\left|\mathrm{\Delta y}\right|$—and communicate the reasoning, units, and rounding in a short written explanation (3–5 sentences) that meets given success criteria.

🧩 Key Ideas & Terms

• Distance (Euclidean) — straight-line length between two points
  $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
• Squared distance — compare using ${\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2$ before taking square roots to avoid rounding errors.
• Triangle classification via squares — with side lengths opposite angles $a$, $b$, $c$ and largest as $c$:
  • Right if $c^2=a^2+b^2$
  • Acute if $c^2<a^2+b^2$
  • Obtuse if $c^2>a^2+b^2$
• City-walk (Manhattan) distance — axis-aligned path length
  $\left|\mathrm{\Delta x}\right|+\left|\mathrm{\Delta y}\right|$
• Largest-square vs. sum test — name the largest squared side as $L$; the sum of the other two as $S$. Compare: right if $L=S$, acute if $L<S$, obtuse if $L>S$.
• Exact form vs. decimal approximation — keep radicals exact (e.g., $5\sqrt{13}$) during work; round only at the end to the required precision. Units and precision — state and use units consistently; follow the specified rounding rule (e.g., nearest tenth of a meter).

🔄 Quick Recall / Prior Knowledge

1. Compute the straight-line distance between $(-2,3)$ and $(4,-5)$. Give exact and decimal (nearest tenth).
   Show Answer

   $\mathrm{\Delta x}=6,\mathrm{\Delta y}=-8;d=\sqrt{6^2+8^2}=\sqrt{100}=10$. Decimal: $10.0$.

2. Simplify exactly: $\sqrt{72}$.
   Show Answer

   $\sqrt{72}=\sqrt{36·2}=6\sqrt{2}$.

3. Classify the triangle with $A=(0,0)$, $B=(6,0)$, $C=(6,8)$. If right, name the vertex of the right angle.
   Show Answer

   Squared lengths: ${\mathrm{AB}}^2=36,{\mathrm{BC}}^2=64,{\mathrm{CA}}^2=100$. Largest is $100$; others sum $36+64=100$. Right triangle; right angle at $B$.

4. From $(3,-2)$ to $(-5,7$ $)$, find Euclidean distance and city-walk distance. Which is larger?
   Show Answer

   Euclidean: $d=\sqrt{{(-8)}^2+9^2}=\sqrt{145}$. City-walk: $|-8|+\left|9\right|=17$. City-walk is larger.

5. If an exact distance is $\sqrt{127}$, report to the nearest tenth.
   Show Answer

   $\sqrt{127}\approx 11.3$ (since it is about 11.269...).

📖 Explore the Lesson

Essential question

How can you use distances on the coordinate plane to solve real situations, decide when a right angle is present, and communicate answers clearly with exact forms, units, and sensible rounding?

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1. Launch: From movement to distance

Imagine a point moving from a start location to a finish location on grid paper. You move horizontally, then vertically. Record the horizontal change and the vertical change. These two numbers are your differences in coordinates: horizontal change is $\mathrm{\Delta x}$, vertical change is $\mathrm{\Delta y}$. The straight-line distance between the two locations is the length of the hypotenuse of a right triangle whose legs have lengths $\left|\mathrm{\Delta x}\right|$ and $\left|\mathrm{\Delta y}\right|$.

The distance formula is therefore:

Show Answer

$d=\sqrt{{\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2}$ where $\mathrm{\Delta x}=x_2-x_1$ and $\mathrm{\Delta y}=y_2-y_1$.

Quick guided try

Find the distance from $(-4,5)$ to $(9,-1)$. Give exact and decimal to the nearest tenth.

Show Answer

Differences: $\mathrm{\Delta x}=13$, $\mathrm{\Delta y}=-6$. Distance: $d=\sqrt{{13}^2+{-6}^2}=\sqrt{205}$. Decimal: $\sqrt{205}\approx 14.3$.

Mini-summary A. Distance on a grid comes from the Pythagorean relationship on the differences. Always compute $\mathrm{\Delta x}$ and $\mathrm{\Delta y}$ first, then square, add, and take a square root only if a final length is required.

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2. Exact first, decimals later

Working exactly keeps comparisons reliable. The square root function can create rounding noise. To avoid that, compare squared distances when your goal is classification or a decision that does not need a rounded length. You may convert to decimals at the end when a measurement must be communicated with a specified precision.

Try it

Without a calculator, compare which is longer: the segment from $(-1,6)$ to $(7,2)$, or the segment from $(3,1)$ to $(-5,10)$.

Show Answer

First segment squared distance: $d^2=8^2+{-4}^2=80$. Second: $d^2={-8}^2+9^2=145$. Since $145>80$, the second segment is longer.

Mini-summary B. When the goal is to compare or classify, keep squares. Take roots at the end only if you must state a length with units and rounding.

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3. The triangle classification test from coordinates

Given three points, you can classify the triangle formed without measuring any angles. Compute the three squared side lengths, identify the largest, then compare it with the sum of the other two. Use the following rule:

• Right if largest equals the sum $c^2=a^2+b^2$
• Acute if largest is less than the sum $c^2<a^2+b^2$
• Obtuse if largest is greater than the sum $c^2>a^2+b^2$

Structured routine for any three points

1. Compute three squared side lengths with a small differences table.
2. Circle the largest squared distance and label it $L$.
3. Add the other two squared distances to get $S$.
4. Compare $L$ with $S$ using the test above.
5. If right, the right angle is at the vertex opposite the side whose squared length is $L$.

Guided example 1

Classify the triangle with $A=(1,1$ $)$, $B=(7,1)$, $C=(7$ $,9$ $)$. If it is right, name the right-angle vertex.

Show Answer

${\mathrm{AB}}^2=6^2+0=36$, ${\mathrm{BC}}^2=0+8^2=64$, ${\mathrm{CA}}^2={-6}^2+{-8}^2=36+64=100$. $L=100$, $S=36+64=100$, so the triangle is right. The right angle is at the vertex opposite side CA, which is point B.

Guided example 2

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

Show Answer

${\mathrm{PQ}}^2={-6}^2+0=36$, ${\mathrm{QR}}^2=0+{-8}^2=64$, ${\mathrm{RP}}^2=6^2+8^2=100$. $L=100$, $S=36+64=100$. The triangle is right, with the right angle at Q.

Mini-summary C. Compute three squared distances, pick the largest, compare with the sum of the other two. Equality gives right. Less gives acute. Greater gives obtuse. The right-angle vertex is the vertex opposite the longest side when equality holds.

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4. Decision making: which distance model fits the situation

In some contexts the path is constrained to horizontal and vertical moves, like a robot in warehouse aisles or a person walking city blocks. Then the relevant distance is the city-walk distance: $\left|\mathrm{\Delta x}\right|+\left|\mathrm{\Delta y}\right|$. In open space, the shortest path is the Euclidean straight line.

Decision check

Identify which model is appropriate in each case and compute the path length.

1. A warehouse robot goes from rack A at $(2,5)$ to rack B at $(17,1)$ along perpendicular aisles only.
2. A drone flies from launch at $(0$$,0)$ to waypoint at $(120,90)$ meters in open air.

Show Answer

1) City-walk model. Distance $|17-2|+|1-5|=15+4=19$ aisle units. 2) Euclidean model. Distance $d=\sqrt{{120}^2+{90}^2}=\sqrt{22500}=150$ meters.

Mini-summary D. Choose the model that matches how movement actually happens. Euclidean works for free movement in open space. City-walk works for axis-aligned routes.

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5. Real-world application set 1: sports, safety, and layout

A training field is marked on a grid where 1 unit represents 1 meter. Cones are placed at $A=(5,2)$, $B=(11,2)$, $C=(11,9)$. A safety requirement says the angle at B must be a right angle.

Task 1. Confirm whether triangle ABC is right and identify the right-angle vertex. Task 2. If a running lane goes straight from A to C, compute its length in meters. Give exact and decimal.

Show Answer

Task 1. Squared distances: ${\mathrm{AB}}^2=6^2+0=36$, ${\mathrm{BC}}^2=0+7^2=49$, ${\mathrm{CA}}^2={-6}^2+{-7}^2=36+49=85$. $L=85$, $S=36+49=85$. Right triangle. Right angle at B. Task 2. Length AC: $d=\sqrt{85}\approx 9.2$ meters.

Mini-summary E. Safety checks that require right angles can be confirmed with squared comparisons. Long diagonals can be reported exactly as radicals, then as decimals with units.

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6. Real-world application set 2: maps and scale drawings

A campus map uses a scale of 1 cm to 50 m. Two buildings appear at coordinates in centimeters $(1,4)$ and $(6$$,10)$.

Task. Compute the real straight-line distance between the buildings. Give exact and then a rounded result to the nearest meter, and explain why rounding happens after scaling.

Show Answer

Map distance in centimeters: $d=\sqrt{{(6-1)}^2+{(10-4)}^2}=\sqrt{5^2+6^2}=\sqrt{61}$ cm. Real distance: multiply by 50 m per cm to get $50\sqrt{61}$ meters. Decimal: $50\sqrt{61}\approx 390.5$ meters, about 391 m. Rounding is done after scaling to keep more precision during multiplication.

Mini-summary F. On scale drawings, compute the geometric distance on the map first, then convert to real units. Keep radicals exact until after conversion, then round.

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7. Real-world application set 3: robots, routes, and energy

A factory robot must move between stations at $(3,2)$ and $(20,17)$ along grid aisles only. The battery estimator needs the actual path length. An engineer also wants to know what the straight-line distance would be for future redesign.

Task. Compute both distances and the extra distance the robot travels compared with the straight line.

Show Answer

City-walk distance: $|20-3|+|17-2|=17+15=32$ aisle units. Euclidean straight-line distance: $\sqrt{{17}^2+{15}^2}=\sqrt{514}$. Extra distance: $32-\sqrt{514}$ units.

Mini-summary G. Choose city-walk for aisle movement because that is the actual path the robot uses. The difference from the Euclidean diagonal shows the cost of constraints.

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8. Method builder: small differences table for speed and accuracy

When triangles are formed by three points, it helps to organize the work in a tiny table before squaring and adding. Fill one row per side.

Example points: $U=(4,4)$, $V=(10,1)$, $W=(1,13)$.

Your turn. Complete the differences table and classify the triangle.

Show Answer

UV: $\mathrm{\Delta x}=6$, $\mathrm{\Delta y}=-3$, squared distance $45$. VW: $\mathrm{\Delta x}=-9$, $\mathrm{\Delta y}=12$, squared distance $225$. WU: $\mathrm{\Delta x}=3$, $\mathrm{\Delta y}=-9$, squared distance $90$. Largest is 225. Sum of other two is 135. Since $225>135$, the triangle is obtuse.

Mini-summary H. A differences table helps prevent sign mistakes and makes the largest-vs-sum comparison fast.

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9. Error clinic: what to avoid and why

Common error 1. Squaring the sum of the differences as if ${(\mathrm{\Delta x}+\mathrm{\Delta y})}^2$ were the same as ${\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2$. This is not valid.

Common error 2. Taking square roots early, rounding, then comparing rounded lengths. This can flip equality into less or greater wrongly.

Common error 3. Dropping signs when computing $\mathrm{\Delta y}$ or $\mathrm{\Delta x}$. Even though squaring clears the sign, the intermediate arithmetic can go wrong if the difference is computed incorrectly.

Repair steps. Always compute differences carefully. Keep squared forms when comparing. Only round at the end when reporting a length.

Quick check. Which of the following is the correct squared distance between $(8,-1)$ and $(-4,7)$: A) ${(-12+8}^2$ B) ${(-12)}^2+8^2$

Show Answer

B) is correct. The squared distance is $144+64=208$. Choice A incorrectly adds differences before squaring.

Mini-summary I. Correct structure matters. Square each difference separately, add, and keep exact squares when you compare.

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10. Investigation: how the squared test matches angle size

Let triangle sides opposite angles be $a$, $b$, $c$ with $c$ the longest. Consider three facts:

1. If the angle opposite $c$ is right, then $c^2=a^2+b^2$.
2. If that angle is acute, then $c^2<a^2+b^2$.
3. If that angle is obtuse, then $c^2>a^2+b^2$.

Check your understanding. If a triangle has squared side lengths 85, 52, and 45, how do you classify it?

Show Answer

Largest is 85. Sum of the other two is 97. Since $85<97$, the triangle is acute.

Mini-summary J. Angle type is encoded in squared lengths. The largest side compared to the sum of the other two tells you right, acute, or obtuse.

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11. Multi-step scenario: plaza cable and utility box

A rectangular plaza has opposite corners at plan coordinates $(2,1)$ and $(15,11)$. The plan scale is 1 cm to 2 m. A diagonal cable will run across the plaza. A utility box is at plan coordinate $(15,1)$.

Tasks. a) Find the plan diagonal length in centimeters. b) Convert to meters and decide the purchase length if cable must be bought in whole meters. c) Form the triangle with the two opposite corners and the utility box. Classify this triangle and identify the right-angle vertex if it is right.

Show Answer

a) Plan diagonal: $d=\sqrt{{(15-2)}^2+{(11}^2}$ $=\sqrt{{13}^2+{10}^2}=\sqrt{269}$ cm. b) Real length: $2\sqrt{269}\approx 32.802$ m. Purchase 33 m. c) Triangle with points C at $(2,1)$, D at $(15,11)$, and B at $(15,1)$. Squared distances: ${\mathrm{CD}}^2=269$, ${\mathrm{DB}}^2=100$, ${\mathrm{BC}}^2=169$. $L=269$, $S=100+169=269$. The triangle is right with right angle at B.

Mini-summary K. A single real scenario can require both a distance computation and a classification. Use exact radicals until the final communication step. Then round and state units.

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12. Near-right sensitivity: why rounding can mislead

Consider points $A=(0,0)$, $B=(9,0)$, $C=(9,12)$. This triangle is right because ${\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{CA}}^2$. Now nudge C to $(8$$,11)$. The largest squared side becomes larger than the sum of the other two and the triangle turns obtuse.

Try it. Prove the new triangle is obtuse by computing the squared distances and comparing.

Show Answer

With C at $(8,11)$: ${\mathrm{AB}}^2=81$, ${\mathrm{BC}}^2=1^{}+{11}^{}=122$, ${\mathrm{CA}}^2=8^{}+{11}^{}=185$. Largest is 185, sum of the other two is 203. Since $185<203$ this choice is actually acute. To force obtuse, move to $(8,13)$: then ${\mathrm{BC}}^2=1^{}+{13}^{}=170$ and ${\mathrm{CA}}^2=233$, with $233>81+170$ false. A simple obtuse example from the original right set is to move to $(10,11)$: then ${\mathrm{BC}}^2=1^{}+{11}^{}=122$ and ${\mathrm{CA}}^2={10}^{}+{11}^{}=221$, with largest 221 and sum 203, so obtuse. The key idea stands: small coordinate moves can flip classification when values are close, which is why rounding early can mislead.

Mini-summary L. Near-equalities are sensitive. Keep squared values exact when classifying. Take square roots only when reporting a final length.

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13. Communication standards: units and precision

When a problem specifies a unit and a rounding rule, follow it. If units are meters and the rule is nearest tenth, present the final distance as, for example, 12.7 m. Do not round during intermediate steps unless instructed.

Practice. A segment has exact length $\sqrt{338}$ meters. Report to the nearest tenth of a meter and explain your rounding step.

Show Answer

$\sqrt{338}\approx 18.384$ meters, which rounds to 18.4 m to the nearest tenth. The hundredth digit is 8 so the tenth rounds up.

Mini-summary M. State units, follow the specified precision, and round only in the final communication step.

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14. Synthesis practice: classify and decide the model

For each item, do all of the following. 1) Compute three squared side lengths. 2) Classify as right, acute, or obtuse. 3) If the situation involves movement along grid aisles, compute the city-walk distance for the segment actually traveled and compare with the Euclidean straight-line segment.

Item A

Points $J=(-2,6)$, $K=(5,6)$, $L=(5,-2)$.

Show Answer

${\mathrm{JK}}^2=7^2+0=49$, ${\mathrm{KL}}^2=0+{-8}^2=64$, ${\mathrm{LJ}}^2={-7}^2+8^2=113$. Right because $49+64=113$. Right angle at K.

Item B

Points $A=(1,2)$, $B=(4,5)$, $C=(7,1)$.

Show Answer

${\mathrm{AB}}^2=18$, ${\mathrm{BC}}^2=20$, ${\mathrm{CA}}^2=52$. Largest 52, sum of others 38, so obtuse.

Item C

Warehouse travel from $(-6,3)$ to $(9,10)$ along aisles only. Compute city-walk distance and the Euclidean distance, then the extra distance.

Show Answer

City-walk: $\left|15\right|+\left|7\right|=22$. Euclidean: $\sqrt{{15}^{}+7^{}}=\sqrt{274}$. Extra: $22-\sqrt{274}$.

Mini-summary N. Classification depends only on the squared values. Movement costs in constrained settings come from city-walk distance.

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15. Design challenge: make a plan with a guaranteed right angle

You must place three stakes on a grid so that the angle at the middle stake is a right angle. You also must route a ribbon along the two legs and then measure the straight diagonal between the first and last stake. Give exact values for the leg lengths and the diagonal. Describe how you guaranteed the right angle.

Your plan. Choose a horizontal segment and a vertical segment that meet at your middle stake.

Show Answer

Example choice: stakes at S $(2,5)$, T $(10,5)$, U $(10,13$$)$. By construction ST is horizontal and TU is vertical, so angle at T is right. Leg lengths are 8 and 8. Diagonal SU has squared length 128, so length $8\sqrt{2}$.

Mini-summary O. To create a right angle on a grid, align one leg horizontally and the other vertically that meet at your chosen vertex.

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16. Evidence and explanation: write to justify your classification

Given three points, your explanation should include a) the three squared side lengths, b) which is largest, c) the comparison statement, d) the conclusion with the vertex if it is right.

Prompt. Explain the classification for points $X=(0,4$$)$, $Y=(9$$,4)$, $Z=(9,15)$.

Show Answer

${\mathrm{XY}}^2=81$, ${\mathrm{YZ}}^2=121$, ${\mathrm{ZX}}^2=202$. Largest is 202, sum of the other two is also 202, so right triangle. The right angle is at Y because it is opposite the longest side ZX.

Mini-summary P. A clear written justification names the three squared values, the largest, the comparison, and the conclusion. If right, it identifies the right-angle vertex.

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17. Mixed applications: construction, navigation, and design

Scenario 1: Construction diagonal

A rectangular frame measures 12 m by 16 m. The diagonal brace must not exceed 20 m. Check the diagonal length and state whether it fits the requirement.

Show Answer

Diagonal length: $\sqrt{{12}^{}+{16}^{}}=\sqrt{400}=20$ m. It meets the at most 20 m limit exactly.

Scenario 2: Navigation segment

From berth A at $(-7,3$$)$ to buoy B at $(9$$,-6)$, find the straight-line distance in nautical miles if 1 grid unit equals 0.5 nautical miles. Give exact and decimal.

Show Answer

Grid distance exact: $\sqrt{{16}^{}+{-9}^{}}=\sqrt{337}$. Real distance: $\frac{1}{2}\sqrt{337}$ nautical miles. Decimal: $\frac{1}{2}\sqrt{337}\approx 9.18$ nautical miles.

Scenario 3: Exhibition design

Three spotlights must form an acute triangle for even coverage. The proposed mounts are at $(2,2)$, $(6$$,7)$, $(10,1)$. Check the classification.

Show Answer

Squared distances: $41$, $52$, $85$. Largest 85, others sum 93, so acute.

Mini-summary Q. Practical designs and routes are solved by the same core processes: compute differences, classify with squared comparisons, choose models that match movement, and present units carefully.

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18. Independent challenge: layered problem with justification

A grid-based delivery path runs from cold storage at $(-3,4$$)$ to packing at $(12$$,9)$. a) If the forklift must move along aisles, what is the path length. b) If the warehouse is redesigned to allow free diagonal movement, what is the straight-line distance. c) A safety triangle is formed with a third point at $(-3,9)$. Classify this triangle and justify fully.

Show Answer

a) City-walk: $|12--3|+|9-4|=15+5=20$ units. b) Euclidean: $\sqrt{{15}^{}+5^{}}=\sqrt{250}=5\sqrt{10}$. c) Triangle with points A $(-3,4)$, B $(12$$,9)$, C $(-3$$,9)$. Squared distances: ${\mathrm{AB}}^2=250$, ${\mathrm{BC}}^2={-15}^{}2+0^{}=225$, ${\mathrm{CA}}^2=0^{}+5^{}=25$. Largest 250, sum of others 250. Right triangle, right angle at C.

Mini-summary R. A layered problem may mix both models and a classification. Tackle each piece in order: path rule, geometric calculation, then classification with squared comparisons.

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19. Self-check statements

Decide true or false and justify briefly in your notebook before opening.

1. The city-walk distance between two points is always at least the Euclidean distance between the same two points.
2. If the largest squared side equals the sum of the other two squared sides, the triangle is right.
3. Rounding square roots before comparing can change a correct classification into an incorrect one.

Show Answer

All three statements are true. Justifications: triangle inequality and path constraints for item 1, converse of the Pythagorean Theorem for item 2, and sensitivity to rounding in near-equalities for item 3.

Mini-summary S. Core truths reinforce your practice. City-walk is never shorter than Euclidean. Equality of squares characterizes right triangles. Rounding is postponed until final reporting.

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20. Reflection prompt for understanding

Write in your notebook: describe a problem where you would choose city-walk distance and another where you would choose Euclidean distance. For each, explain in 3 to 5 sentences how you would compute and how you would present your final answer with units and rounding.

Show Answer

Example answers: City-walk for an indoor robot that moves along aisles. Compute $\left|\mathrm{\Delta x}\right|+\left|\mathrm{\Delta y}\right|$, then state aisle units or meters and any rounding direction. Euclidean for a drone or line-of-sight cable. Compute $\sqrt{{\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2}$, keep exact until the end, then round with units.

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21. Final checkpoint: quick mixed items

Q1. Distance from $(-5,2$$)$ to $(7$$,-9)$. Exact and decimal to nearest tenth. Q2. Classify triangle with points $M=(0$$,0)$, $N=(8$$,6)$, $P=(0$$,6)$. If right, name the right-angle vertex. Q3. City-walk vs Euclidean from $(4$$,-3)$ to $(-10$$,5)$. Give both and the extra distance.

Show Answer

Q1. Differences 12 and -11. Exact length $\sqrt{265}$, decimal about 16.3. Q2. Squared distances 100, 64, 36. Since $64+36=100$, right triangle, right angle at N. Q3. City-walk $14+8$$=22$. Euclidean $\sqrt{260}$. Extra $22-\sqrt{260}$.

Mini-summary T. You can now handle mixed tasks that require careful model choice, exact comparisons, and clear communication of lengths with units.

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References

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

What you can now do: compute straight-line distances exactly and approximately; classify triangles by comparing the largest squared side with the sum of the other two and identify the right-angle vertex when applicable; choose between Euclidean and city-walk models to match real movement and present answers with appropriate units and rounding.

💡 Example in Action

Example 1 — Straight-line distance with exact and decimal forms

Find the distance between the points $(-7,4)$ and $(8$$,-3)$. Report the exact value and the decimal to the nearest tenth (units).

Show Answer

Step 1. Compute differences. $\mathrm{\Delta x}=15,\mathrm{\Delta y}=-7$.
Step 2. Distance (exact). $d=\sqrt{{15}^{}+{-7}^{}}=\sqrt{274}$.
Step 3. Decimal. $\sqrt{274}\approx 16.6$.
Answer. Exact $\sqrt{274}$; decimal $16.6$ units.

Example 2 — Classify a triangle from three coordinates

Classify the triangle formed by $A=(2$$,5)$, $B=(10$$,5)$, $C=(6$$,13)$. If it is right, identify the right-angle vertex.

Show Answer

AB: ${\mathrm{AB}}^2=64$; BC: ${\mathrm{BC}}^2=80$; CA: ${\mathrm{CA}}^2=80$. Largest $L=80$, sum of others $S=144$; $L<S$ so acute.

Example 3 — Choose the distance model and quantify the “extra”

A cart must travel along grid aisles from $(3$$,-2)$ to $(-9$$,9)$. Find the city-walk distance, the Euclidean straight-line distance, and the extra distance.

Show Answer

City-walk $23$ units. Euclidean $\sqrt{265}$ (≈ $16.3$). Extra $23-\sqrt{265}$.

Example 4 — Scale map to real distance with proper rounding

On a plan where 1 cm represents 25 m, two locations are at $(1.4,3.8)$ cm and $(5.9$$,-2.2)$ cm.

Show Answer

Map distance $7.5$ cm; real distance $187.5$ m; rounded $188$ m.

Example 5 — Presenting with units and a chosen precision

A surveyor measures from $(0$$,0)$ to $(1200$$,900)$ in meters. Report the straight-line distance in kilometers to the nearest hundredth.

Show Answer

Distance $1500$ m; in km $1.50$ km.

📝 Try It Out

1. Distance between $(-6,5)$ and $(9$$,-3)$.
   Show Answer

   Exact $17$; decimal $17.0$.

2. City-walk vs Euclidean from $(3$$,-4)$ to $(-8$$,9)$.
   Show Answer

   City-walk $25$; Euclidean $\sqrt{290}$ ≈ $17.0$; difference $25-\sqrt{290}$.

3. Classify triangle with A(0,0), B(8,0), C(5,12).
   Show Answer

   Acute; squares 64, 153, 169 with 169 < 217.

4. Compare two segments without roots.
   Show Answer

   S₂ longer (squared 180 vs 72).

5. Scale map to real distance.
   Show Answer

   Exact $\sqrt{78.73}$ cm; real ≈ $355$ m.

6. Choose model and extra.
   Show Answer

   City-walk 19; Euclidean $\sqrt{185}$ ≈ 13.6; extra ≈ 5.4.

7. Determine $k$ for right at C.
   Show Answer

   $k=6$ (non-degenerate).

8. Rounding $5\sqrt{13}$ m.
   Show Answer

   ≈ $18.0$ m.

9. Classify triangle PQR.
   Show Answer

   Acute (160 < 170).

10. Longer from origin.
    Show Answer

    (12,5) longer (169 > 162).

✅ Check Yourself

1. Distance between (−3,7) and (9,−2).
   Show Answer

   Exact $15$; decimal $15.0$.

2. Triangle with A(0,0), B(9,0), C(9,12): classify and right-angle vertex.
   Show Answer

   Right; vertex B.

3. Warehouse travel from (2,−5) to (−8,6).
   Show Answer

   City-walk 21; Euclidean $\sqrt{221}$ ≈ 14.9; extra ≈ 6.1.

4. Scale 1 cm = 30 m, points (1.2,3.5) and (7.8,−2.1).
   Show Answer

   Real distance $6\sqrt{1873}$ m ≈ 260 m.

5. Longer segment without roots.
   Show Answer

   S₂ longer (113 vs 90).

6. Find $k$ for right at B with C(k,10).
   Show Answer

   $k=4.5$.

7. Distance between (−4,9) and (2,−3).
   Show Answer

   $6\sqrt{5}$ ≈ 13.4.

8. Classify A(−1,7), B(5,2), C(9,10).
   Show Answer

   Acute.

9. From (−8,−4) to (3,2): distances.
   Show Answer

   City-walk 17; Euclidean $\sqrt{157}$ ≈ 12.5; extra ≈ 4.5.

10. Construct right angle with D on x=3.
    Show Answer

    D(3,3) or D(3,−3).

11. Rectangle 14×48: diagonal vs 50 m.
    Show Answer

    Exactly 50 m; not longer.

12. Cable from (−12,5) to (8,−15): nearest tenth.
    Show Answer

    $20\sqrt{2}$ ≈ 28.3 m.

13. Classify S(−4,1), T(2,−11), U(10,7).
    Show Answer

    Acute.

14. Concept check: City-walk equals Euclidean iff one difference is zero.
    Show Answer

    True.

15. Right-angle vertex and perimeter for A(−1,−1), B(7,−1), C(7,10).
    Show Answer

    Right at B. Perimeter $8+11+\sqrt{185}$ ≈ 32.6.

🚀 Go Further

1) Integer-coordinate right triangle with integer side lengths

Show Answer

Use 5–12–13 with A(0,0), B(5,0), C(5,12). Lengths 5,12,13.

2) Locus of all points forming a right triangle with fixed endpoints

Show Answer

Circle with diameter AC; center (2,3), radius squared 20; equation ${(x-2)}^2+{(y-3)}^2=20$.

3) Maximize the “grid penalty” (city-walk minus Euclidean)

Show Answer

Maximum $40-20\sqrt{2}$ for |Δx|=|Δy|=20.

4) Prove the city-walk distance is never shorter than Euclidean

Show Answer

Square both sides: extra cross term $2|\mathrm{\Delta x}·\mathrm{\Delta y}|$ ≥ 0. Equality iff $\mathrm{\Delta x}=0$ or $\mathrm{\Delta y}=0$.

5) Parameter exploration: when is the triangle acute, right, or obtuse?

Show Answer

With A(0,0), B(6,0), C(t,8): right at t=6; obtuse for t>6; acute for t<6 (angle at B).

🔗 My Reflection

Instructions: In your notebook, complete a 3–2–1 Reflection about today’s work on distances and triangle classification. Be concrete and cite at least one example from your own solutions. Time: 8–10 minutes.

3–2–1 Reflection (write in your notebook)

3 — I can… Write three brief statements describing skills you can now do reliably (use precise math words).

2 — I’m still unsure about… List two points of confusion; turn each into a question.

1 — Next-step action Choose one concrete action you’ll try next time.

Evidence (attach in your notebook): Pick one problem you solved today and justify it briefly (differences, squared values or units/rounding, and model choice).

Success check (tick in your notebook):

• I stated exact values first; I rounded only at the end.
• I used units correctly when required.
• My justification named the largest squared side and compared it to the sum of the other two (for classification), or explained the model choice (for distance).

Sample Reflection (model only—use your own work)

3 — I can…
1) I can compute Euclidean distance by finding $\mathrm{\Delta x}$ and $\mathrm{\Delta y}$, then using $d=\sqrt{{\mathrm{\Delta x}}^2+{\mathrm{\Delta y}}^2}$.
2) I can classify triangles by squares: right if $L=S$, acute if $L<S$, obtuse if $L>S$.
3) I can choose city-walk distance $\left|\mathrm{\Delta x}\right|+\left|\mathrm{\Delta y}\right|$ when movement is along aisles.

2 — I’m still unsure about…
• When to simplify radicals vs. leave them as-is.
• How close values can be before rounding risks a wrong classification.

1 — Next-step action
I will keep squared values for classification and round only in the final reported length with units.

Evidence (from my work):
From a triangle with points $J=(-2,6)$, $K=(5,6)$, $L=(5,-2)$, I found ${\mathrm{JK}}^2=49$, ${\mathrm{KL}}^2=64$, ${\mathrm{LJ}}^2=113$. Since $49+64=113$, the triangle is right with right angle at K. This shows correct use of the largest-square vs. sum test.