MAT8 Q2W5D2: Understanding and Applying the Pythagorean Theorem

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

1. Use the Pythagorean Theorem to find a missing side in right triangles arising from real contexts (including non-integer results), and justify each step.
2. Derive and write the distance formula from the Pythagorean Theorem and compute the straight-line distance between two points on the Cartesian plane.
3. Interpret and evaluate diagonal or straight-line distances to solve design, navigation, and layout problems (for example, maps, fields, plans).

• Cartesian plane - the coordinate plane with perpendicular axes.
• Coordinates - ordered pair $(x_1,y_1)$, position of a point.
• Distance formula - derived from Pythagorean Theorem:
  $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
• Euclidean distance - straight-line distance between two points.
• Pythagorean triple - integers $(a,b,c)$ with $c^2=a^2+b^2$ (for example, 3-4-5, 5-12-13).

1) Identify the hypotenuse.
A right triangle has sides labeled $a$, $b$, and $c$, where the right angle is between sides $a$ and $b$. Which side is the hypotenuse?

Show Answer

c is the hypotenuse (it is opposite the right angle and the longest side).

2) Find a missing leg.
Given hypotenuse $c=13$ and leg $b=5$, find $a$.

Show Answer

$a^2=c^2-b^2$ $\Rightarrow$ $a^2=169-25=144$ $\Rightarrow$ $a=12$.

3) True or false.
If $c^2=a^2+b^2$, the triangle is right.

Show Answer

True. This is the converse of the Pythagorean Theorem.

A. Launch: From Day 1 Triangles to Today’s Coordinates

Yesterday, you explored right triangles: two legs meeting at a right angle and a hypotenuse opposite the right angle. You discovered and used:

$c^2=a^2+b^2$

Today, you will transfer that relationship to the coordinate plane, deriving and applying the distance formula to compute straight-line distances between two points. You will also justify your steps and interpret answers in real settings like maps, sports fields, and basic design plans.

Guiding Question A1
Why might the Pythagorean Theorem help with distances on a coordinate grid?

Show Answer

Because two points and the horizontal or vertical grid lines form a right triangle: the horizontal change and vertical change are the legs, and the straight-line distance between the points is the hypotenuse.

Mini-Summary A
A coordinate grid naturally forms right triangles between any two points by using horizontal and vertical moves. That is why the Pythagorean Theorem can be used to find the straight-line distance.

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B. Building the Bridge: Plot, Drop, and Derive

Consider two points on the plane:

• Point 1: $(x_1,y_1)$
• Point 2: $(x_2,y_2)$

If you draw a horizontal segment from Point 1 to directly under or over Point 2, its length equals the change in x. If you draw a vertical segment from there to Point 2, its length equals the change in y. These two segments are perpendicular, forming a right triangle whose hypotenuse is the straight segment connecting the two points.

Let the horizontal leg be: $|x_2-x_1|$ and the vertical leg be: $|y_2-y_1|$. Since these lengths are squared in the theorem, the absolute value bars are not necessary inside the squares (negative differences square to positive values).

Applying Pythagoras with legs equal to the horizontal and vertical changes, and hypotenuse equal to the straight-line distance $d$:

$d^2={(x_2-x_1)}^2+{(y_2-y_1)}^2$

Taking the positive square root (distance is nonnegative):

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Guiding Question B1
Why do not we need absolute value bars inside the radicals for the differences?

Show Answer

Because we square the differences: ${(-5)}^2=25$, which is the same as ${\left(5\right)}^2=25$. Squaring removes the sign.

Mini-Summary B
The distance formula is simply the Pythagorean Theorem applied to horizontal and vertical changes between two points.

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C. First Contact: Compute a Distance on the Grid

Activity C1 – Plot and Measure
Plot the points $(2,5)$ and $(8,1)$ on grid paper. Draw the right triangle by dropping a horizontal and a vertical segment.

1. What is the horizontal change $x_2-x_1$?
2. What is the vertical change $y_2-y_1$?
3. Compute the distance using the formula.

Show Answer

1) Horizontal change: $x_2-x_1=8-2=6$

2) Vertical change: $y_2-y_1=1-5=-4$

3) Distance: $d=\sqrt{6^2+{(-4)}^2}=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}\approx 7.21$

Guiding Question C2
If you swapped the order of subtraction for the x’s or y’s, would the distance change?

Show Answer

No. Although the difference might switch sign, it is squared, so the final distance is unchanged.

Mini-Summary C
Distance depends on the sizes of the horizontal and vertical changes, not on their signs.

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D. From Formula to Sense-Making: Units, Scale, and Rounding

In the coordinate plane, each grid step usually represents one unit. In real maps, one grid step could represent a real distance (like 1 cm = 1 km). Distances should be reported with appropriate units and appropriate precision.

Guiding Question D1
Why is it often better to keep an exact square-root form before giving a rounded decimal?

Show Answer

The exact form (like $\sqrt{52}$ or $2\sqrt{13}$) preserves full precision. Decimal rounding can introduce small errors, so we keep the radical form until the final step or until a specified precision is required.

Mini-Summary D
Keep exact forms as long as possible. When interpreting results in the real world, include units and round reasonably.

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E. City Blocks vs. Straight Lines: Shortest Paths and Constraints

On a grid city with only east or west and north or south streets, you usually cannot travel diagonally. In that case, your travel distance along streets is different from the straight-line distance the formula gives.

Activity E1 – City Walk
You start at $(1,2)$ and need to reach $(9,10)$.

1. Compute the Euclidean (straight-line) distance.
2. If you can only move horizontally and vertically, what is the minimum city-walking distance?

Show Answer

1) Straight-line (Euclidean) distance: $d=\sqrt{{(9-1)}^2+{(10-2)}^2}=\sqrt{8^2+8^2}=\sqrt{64+64}=\sqrt{128}=8\sqrt{2}\approx 11.31$

2) City-walking distance (no diagonals): total horizontal plus total vertical: $|9-1|+|10-2|=8+8=16$

Guiding Question E2
Which distance better represents the physical straightness of the path? Which better represents actual travel in a grid city with no diagonal streets?

Show Answer

The Euclidean distance represents the straight line in open space; the city-walking sum represents actual travel if you are constrained to horizontal or vertical streets.

Mini-Summary E
Models matter. The distance formula gives straight-line distance; real travel may need a different model.

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F. Deriving the Distance Formula Again (Geometric Reasoning)

It is helpful to reason visually: draw a rectangle whose opposite corners are your two points. The rectangle’s width is $|x_2-x_1|$, and height is $|y_2-y_1|$. The diagonal of that rectangle is exactly the distance you want, so Pythagoras applies directly.

Guiding Question F1
How do you know the angle between the horizontal and vertical legs is 90°?

Show Answer

Horizontal and vertical directions are perpendicular by definition in the Cartesian plane, so they form a right angle.

Mini-Summary F
The coordinate plane guarantees right angles between horizontal and vertical directions; that is the backbone of the distance formula.

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G. Worked Walkthroughs

We will compute distances with integers, negatives, and mixed coordinates. Focus on clear steps and justifications.

Example G1 – Integer Coordinates
Find the distance between $(-3,4)$ and $(5,-2)$.

Show Answer

Compute differences: $x_2-x_1=5--3=8$, $y_2-y_1=-2-4=-6$

Distance: $d=\sqrt{8^2+{(-6)}^2}=\sqrt{64+36}=\sqrt{100}=10$

Guiding Question G1a
Why is it valid to treat left or down movement as negative differences?

Show Answer

Because direction is captured by sign, and the formula squares the differences, making the final contribution to distance nonnegative.

Example G2 – Fractional or Decimal Coordinates
Find the distance between $(2.5,-1.2)$ and $(-3.1,2.8)$.

Show Answer

Differences: $x_2-x_1=-3.1-2.5=-5.6$, $y_2-y_1=2.8--1.2=4.0$

Distance: $d=\sqrt{{(-5.6)}^2+{4.0}^2}=\sqrt{31.36+16}=\sqrt{47.36}=6.88$

Guiding Question G2a
Why might a designer prefer to keep values in exact fractional form during intermediate steps?

Show Answer

To avoid rounding errors accumulating; exact forms preserve precision until the final answer is required.

Mini-Summary G
The distance formula handles integers, negatives, fractions, and decimals the same way: by squaring and adding the horizontal and vertical changes.

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H. Diagonals in Design: Layouts, Screens, and Fields

Activity H1 – Screen Diagonal
A monitor’s width is $48$ cm and its height is $27$ cm. The diagonal is the straight-line distance across the rectangle.

1. Model the diagonal with a right triangle – what are the legs?
2. Compute the diagonal length (exact and to the nearest tenth).

Show Answer

1) Legs: $48$ and $27$.

2) Diagonal: $d=\sqrt{{48}^2+{27}^2}=\sqrt{2304+729}=\sqrt{3033}\approx 55.1$

Guiding Question H2
If a screen is advertised as 55-inch, what measurement does that refer to?

Show Answer

The diagonal length.

Mini-Summary H
Rectangles hide right triangles in their diagonals. The distance formula and Pythagoras both retrieve the diagonal length.

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I. Navigation and GPS: Straight vs. Planned Path

Imagine a drone takes off from a point A and flies to point B. The drone’s navigation system uses coordinates relative to its starting point.

• A = $(0,0)$
• B = $(120,90)$ (meters)

Activity I1 – Drone Hop

1. Compute the straight-line distance from A to B.
2. If wind forces the drone to move in two legs – first east the full horizontal amount, then north the full vertical amount – how much farther is that compared to the straight-line path?

Show Answer

1) Straight-line: $d=\sqrt{{120}^2+{90}^2}=\sqrt{14400+8100}=\sqrt{22500}=150$ $\text{meters}$

2) Two-leg path: $120$ + $90$ = $210$ m. Extra distance = $210-150=60$ m.

Guiding Question I2
Why do pilots, bikers, or hikers sometimes prefer a path that is not the Euclidean straight line?

Show Answer

Obstacles, terrain, regulations, wind, roads, or safety concerns might constrain the path, making a non-straight route more practical or even mandatory.

Mini-Summary I
Euclidean distance is the ideal straight line; real travel considers constraints and safety.

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J. Classroom Lab: Create Right Triangles from Coordinates

Activity J1 – Make and Measure
Pick any two points:

• P: $(-4,7)$
• Q: $(5,-3)$

1. Sketch and draw the right triangle using horizontal or vertical moves.
2. Compute the run and rise.
3. Compute the distance $d$.

Show Answer

2) Run: $x_2-x_1=5--4=9$

Rise: $y_2-y_1=-3-7=-10$

3) Distance: $d=\sqrt{9^2+{(-10)}^2}=\sqrt{81+100}=\sqrt{181}\approx 13.45$

Guiding Question J2
How can you check quickly if you have made a sign error in differences?

Show Answer

Sketch the right triangle: the lengths of the legs must be nonnegative. If your squared values or totals look wrong (like a negative under a square root), re-check signs and arithmetic.

Mini-Summary J
Sketching is a powerful error-check. The geometry gives intuition to verify the algebra.

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K. Interpreting Answers: Units, Context, and Reasonableness

• Units: If your coordinates are meters, the answer is in meters.
• Scale: On a map with a scale, convert grid or cm to real distance.
• Reasonableness: Does the distance fit the drawing? Is the value too big or too small?

Activity K1 – Scale Check
A campus map uses a 1 cm grid where 1 cm represents 50 m in real life. Two buildings are at $(1,4)$ and $(6,10)$ on the map grid. Find the real straight-line distance.

Show Answer

Map distance in cm: $d=\sqrt{{(6-1)}^2+{(10-4)}^2}=\sqrt{5^2+6^2}=\sqrt{25+36}=\sqrt{61}\approx 7.81$ $\text{cm}$

Real distance: multiply by 50 m per cm $7.81\times 50=390.5$ $\text{m (≈ 391 m)}$

Mini-Summary K
Mind the scale and units. The same formula supports both simple grids and real-world maps.

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L. Pythagorean Triples in Coordinates

If the horizontal and vertical differences form a known Pythagorean triple, the distance becomes a whole number.

Activity L1 – Quick Triple
Compute the distance between $(2,3)$ and $(-7,9)$.

Show Answer

Horizontal difference: $-7-2=-9$ (length 9)

Vertical difference: $9-3=6$

Distance: $d=\sqrt{9^2+6^2}=\sqrt{81+36}=\sqrt{117}$ $\text{(not triple 3-4-5 scaled, but note 9:6 = 3:2 is not from a common triple).}$

If the leg lengths were 9 and 12, we would have $\sqrt{81+144}=\sqrt{225}=15$, a 3-4-5 scaled triple.

Guiding Question L2
How can recognizing triples speed up your mental calculations?

Show Answer

If the leg lengths match a known triple (or scale of one), you can jump straight to the integer hypotenuse, saving time.

Mini-Summary L
Recognizing triples turns some radical distances into whole numbers instantly.

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M. Design Case: Rectangular Field Planning

A school is planning a rectangular activity field 60 m by 80 m. The engineer needs the diagonal for a lighting cable.

Activity M1 – Field Diagonal
Compute the diagonal and interpret it in context.

Show Answer

$d=\sqrt{{60}^2+{80}^2}=\sqrt{3600+6400}=\sqrt{10000}=100$ $\text{m. So the straight cable across the field should be 100 m (plus slack or tolerances in practice).}$

Mini-Summary M
Field, court, and room diagonals are routine uses of Pythagoras and the distance formula.

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N. Error-Spotting: Common Pitfalls

1. Forgetting to square each difference.
2. Adding differences first, then squaring the total (incorrect).
3. Dropping units in real-world problems.
4. Rounding too early, causing large final errors.

Guiding Question N1
Which is correct:

- $\sqrt{(x_2-x_1)+(y_2-y_1)}$

or

- $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Explain.

Show Answer

The second is correct: each difference must be squared and added, then square-rooted.

Mini-Summary N
Always square differences before adding; keep units; round at the end.

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O. Synthesis: Why This Matters

• Design (diagonals, placements, clearances)
• Navigation (shortest routes, range checks)
• Sports or PE (diagonal sprints, field layouts)
• STEM projects (drones, robots, surveying)

Guiding Question O1
Describe a problem from your life or school where you could use the distance formula.

Show Answer

Answers will vary – examples: measuring a diagonal banner across a stage, estimating the straight-line distance between two campus buildings, or checking if a garden hose reaches a far corner.

Mini-Summary O
You have taken the Pythagorean Theorem beyond triangles – into coordinates, designs, and decisions.

References

• Euclid. Elements (Book I).
• Posamentier, A. and Lehmann, I. The Secrets of Triangles.
• Maor, E. The Pythagorean Theorem: A 4,000-Year History.
• OpenStax. Geometry (Distance on the Coordinate Plane).
• Khan Academy. Distance Formula (concept and practice).

Worked Example 1 – Distance between two points (integer coordinates)

Find the distance between points $(2,-3)$ and $(-4,5)$.

Show Answer

Step 1 – Recall distance formula.

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Step 2 – Substitute differences.

$x_2-x_1=-4-2=-6$ , $y_2-y_1=5--3=8$

Step 3 – Compute.

$d=\sqrt{{(-6)}^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$

Answer: $10$

Worked Example 2 – Distance (decimal coordinates)

Find the distance between $(1.8,-2.4)$ and $(-4.7,3.2)$.

Show Answer

Step 1 – Differences.

$x_2-x_1=-4.7-1.8=-6.5$ , $y_2-y_1=3.2--2.4=5.6$

Step 2 – Distance.

$d=\sqrt{{(-6.5)}^2+{5.6}^2}=\sqrt{42.25+31.36}=\sqrt{73.61}\approx 8.58$

Answer: exact $\sqrt{73.61}$, approximately $8.58$.

Worked Example 3 – Map scale conversion

On a map, points are at $(3,1)$ and $(10,5)$ (in centimeters). Scale: $1\text{cm}=0.5\text{km}$. Find the real straight-line distance.

Show Answer

Step 1 – Map distance (cm).

$d=\sqrt{{(10-3)}^2+{(5-1)}^2}=\sqrt{7^2+4^2}=\sqrt{49+16}=\sqrt{65}$

Step 2 – Convert to kilometers.

$\text{Real distance}=0.5\times \sqrt{65}\approx 0.5\times 8.0623=4.03115$ $\text{km}$

Answer: exact $\frac{1}{2}\sqrt{65}$ km, approximately $4.03$ km.

Worked Example 4 – Find a missing coordinate given a distance

Point A is $(2,-1)$. Point B is $(x,7)$. If the distance $\mathrm{AB}=10$, find all possible values of $x$.

Show Answer

Step 1 – Set up equation.

${10}^2={(x-2)}^2+{(7--1)}^2$

Step 2 – Simplify.

$100={(x-2)}^2+8^2={(x-2)}^2+64$

Step 3 – Solve.

${(x-2)}^2=36\Rightarrow x-2=\pm 6$

Solutions: $x=8$ or $x=-4$.

Worked Example 5 – Check if three points form a right triangle

Do points $(2,3)$, $(2,-1)$, and $(11,-1)$ form a right triangle?

Show Answer

Compute squared lengths.

$\text{AB}:{(2-2)}^2+{(-1-3)}^2=0+{(-4)}^2=16$

$\text{BC}:{(11-2)}^2+{(-1--1)}^2=9^2+0^2=81$

$\text{AC}:{(11-2)}^2+{(-1-3)}^2=9^2+{(-4)}^2=97$

Check Pythagorean relation.

$16+81=97$

Conclusion: Yes, it is a right triangle (right angle at the middle point $(2,-1)$).

Now You Try

1. Find the distance between $(-2,5)$ and $(4,-1)$.
   Show Answer

   $d=\sqrt{{(4--2)}^2+{(-1-5)}^2}=\sqrt{6^2+{(-6)}^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\approx 8.49$

2. Find the distance between $(3.4,-2.1)$ and $(-1.6,4.9)$.
   Show Answer

   $d=\sqrt{{(-1.6-3.4)}^2+{(4.9--2.1)}^2}=\sqrt{{(-5.0)}^2+{7.0}^2}=\sqrt{25+49}=\sqrt{74}\approx 8.60$

3. Map coordinates (in cm): $(1,2)$ and $(7,11)$. Scale: $1\text{cm}=2\text{km}$. Find the real straight-line distance.
   Show Answer

   $\text{Map distance}=\sqrt{{(7-1)}^2+{(11-2)}^2}=\sqrt{6^2+9^2}=\sqrt{36+81}=\sqrt{117}$

   $\text{Real distance}=2\times \sqrt{117}\approx 2\times 10.8167=21.6334$ $\text{km ≈}$ $21.63$$\text{km}$

4. Let A be $(-3,2)$ and B be $(5,y)$. If $\mathrm{AB}=10$, find $y$.
   Show Answer

   ${10}^2={(5--3)}^2+{(y-2)}^2\Rightarrow 100=8^2+{(y-2)}^2\Rightarrow {(y-2)}^2=36$

   Solutions: $y=8$ or $y=-4$.

5. Do points $(0,0)$, $(9,0)$, and $(9,12)$ form a right triangle?
   Show Answer

   $\text{AB}:9^2=81$ $,$ $\text{BC}:{12}^2=144$ $,$ $\text{AC}:{15}^2=225$

   $81+144=225$

   Conclusion: Yes, right triangle (right angle at $(9,0)$).

1. Find the distance between $(-2,7)$ and $(4,-1)$.
   Show Answer

   $d=\sqrt{{(4--2)}^2+{(-1-7)}^2}=\sqrt{6^2+{(-8)}^2}=\sqrt{36+64}=\sqrt{100}=10$

2. Find the distance between $(3.5,-2)$ and $(-1.5,4)$.
   Show Answer

   $d=\sqrt{{(-1.5-3.5)}^2+{(4--2)}^2}=\sqrt{{(-5.0)}^2+6^2}=\sqrt{25+36}=\sqrt{61}\approx 7.81$

3. Map points (in cm) are $(1,4)$ and $(9,10)$. Scale: $1\text{cm}=0.8\text{km}$. Find the real straight-line distance.
   Show Answer

   $\text{Map distance}=\sqrt{{(9-1)}^2+{(10-4)}^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$ $\text{cm}$

   $\text{Real distance}=10\times 0.8=8$ $\text{km}$

4. Let A be $(-3,5)$ and B be $(x,-1)$. If $\mathrm{AB}=10$, find $x$.
   Show Answer

   ${10}^2={(x--3)}^2+{(-1-5)}^2\Rightarrow 100={(x+3)}^2+{(-6)}^2={(x+3)}^2+36$

   ${(x+3)}^2=64\Rightarrow x+3=\pm 8$

   Solutions: $x=5$ or $x=-11$.

5. Do the points $(0,0)$, $(6,8)$, and $(6,0)$ form a right triangle?
   Show Answer

   $\text{AB}:6^2+8^2=100$, $\text{BC}:8^2=64$, $\text{AC}:6^2=36$

   $64+36=100$

   Yes, it is a right triangle (right angle at $(6,0)$).

6. Find the distance between $(-7,-3)$ and $(-1,9)$.
   Show Answer

   $d=\sqrt{{(-1--7)}^2+{(9--3)}^2}=\sqrt{6^2+{12}^2}=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}\approx 13.42$

7. A rectangular park is 120 m by 50 m. Find the diagonal walkway length.
   Show Answer

   $d=\sqrt{{120}^2+{50}^2}=\sqrt{14400+2500}=\sqrt{16900}=130$ $\text{m}$

8. Find the distance between $(2,-3)$ and $(-5,4)$.
   Show Answer

   $d=\sqrt{{(-5-2)}^2+{(4--3)}^2}=\sqrt{{(-7)}^2+7^2}=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}\approx 9.90$

9. Find all values of $y$ if the distance between $(2,y)$ and $(-1,8)$ is $10$.
   Show Answer

   ${10}^2={(2--1)}^2+{(y-8)}^2\Rightarrow 100=3^2+{(y-8)}^2\Rightarrow {(y-8)}^2=91$

   Solutions: $y=8+\sqrt{91}$ or $y=8-\sqrt{91}$.

10. A point moves from $(-2,4)$ by going 9 units east and 12 units south, ending at $(7,-8)$. Find the straight-line distance from start to end.
    Show Answer

    $d=\sqrt{{(7--2)}^2+{(-8-4)}^2}=\sqrt{9^2+{(-12)}^2}=\sqrt{81+144}=\sqrt{225}=15$

1. State the distance formula.
   Show Answer

   $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

2. Concept: Why are not absolute value bars needed inside the squares?
   Show Answer

   Squaring removes sign: ${(-k)}^2=k^2$.

3. Distance between points: $(-4,7)$ and $(9,-1)$.
   Show Answer

   $d=\sqrt{{(9--4)}^2+{(-1-7)}^2}=\sqrt{{13}^2+{(-8)}^2}=\sqrt{233}\approx 15.26$

4. Map scale: Points (in cm) are $(2,1)$ and $(9,5)$. Scale: $1\text{cm}=250\text{m}$. Find real distance.
   Show Answer

   Map distance: $\sqrt{7^2+4^2}=\sqrt{65}\approx 8.0623$ $\text{cm.}$ $\text{Real}=250\times \sqrt{65}\approx 2015.6\text{m}=2.016\text{km (approx.)}$

5. Find $x$ if distance between A $(-5,2)$ and B $(x,-10)$ is $13$.
   Show Answer

   ${13}^2={(x--5)}^2+{(-10-2)}^2\Rightarrow 169={(x+5)}^2+144$

   ${(x+5)}^2=25\Rightarrow x+5=\pm 5$ $\text{so}$ $x=0$ $\text{or}$ $x=-10$.

6. Do the points $(0,0)$, $(0,12)$, and $(5,0)$ form a right triangle?
   Show Answer

   Lengths: $12$, $5$, $13$. Check: $5^2+{12}^2={13}^2$. Yes.

7. City-walk vs straight-line from $(1,2)$ to $(10,14)$.
   Show Answer

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

8. Distance between $(-1.2,3.5)$ and $(4.8,-2.5)$.
   Show Answer

   Differences: $6.0$, $-6.0$. Distance: $\sqrt{36+36}=\sqrt{72}=6\sqrt{2}\approx 8.49$.

9. A design requires the diagonal to be at most 20 m. Rectangle: 12 m by 16 m. Is it acceptable?
   Show Answer

   Diagonal: $\sqrt{144+256}=\sqrt{400}=20\text{m}$. Meets the $\le$ 20 m requirement (borderline).

10. Ladder: Length 25 ft, reaches 24 ft high. How far from the wall is the base?
    Show Answer

    $\sqrt{625-576}=\sqrt{49}=7\text{ft}$.

11. Find all $y$ if distance between $(7,y)$ and $(-1,-2)$ is 10.
    Show Answer

    ${10}^2={(-1-7)}^2+{(-2-y)}^2$ $\Rightarrow$ $100=64+{(-2-y)}^2$ $\Rightarrow$ ${(y+2)}^2=36$. Thus $y=4$ or $y=-8$.

12. Distance between $(-6,-8)$ and $(3,4)$.
    Show Answer

    Differences: $9$, $12$. Distance: $\sqrt{81+144}=15$.

13. Rounding: Compute $d=2\sqrt{13}$ to the nearest tenth.
    Show Answer

    $\sqrt{13}\approx 3.606$, so $d\approx 7.212$ $\text{→ nearest tenth}$ $7.2$.

14. Recognize a triple: Distance between $(5,8)$ and $(14,20)$.
    Show Answer

    Differences: $9$, $12$ $\text{→}$ 3-4-5 scaled. Distance $15$.

15. True or False: Swapping $(x_1,y_1)$ and $(x_2,y_2)$ changes the distance value.
    Show Answer

    False. Differences change sign but are squared: ${(x_2-x_1)}^2={(x_1-x_2)}^2$.

Extension 1 – Straight-line vs. city-walk (inequality)

In a grid city, the city-walk distance between two points is $|x_2-x_1|+|y_2-y_1|$, while the straight-line (Euclidean) distance is $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$. Show that the Euclidean distance is always less than or equal to the city-walk distance, and identify when equality holds.

Show Guidance/Answer

Idea: Let $a$ and $b$ be the nonnegative leg lengths (horizontal and vertical changes). Compare $\sqrt{a^2+b^2}$ with $a+b$.

Work: Square both sides of the desired inequality (all quantities are nonnegative):

${(a+b)}^2=a^2+b^2+2ab\ge a^2+b^2$

Thus $a+b$ is at least $\sqrt{a^2+b^2}$. Equality holds only when $ab=0$, meaning one leg is zero (purely horizontal or purely vertical move).

Extension 2 – Locus of points at a fixed distance (circle)

Find the set of all points at distance $5$ from the center $(3,-2)$. Write the equation and decide if the point $(7,-2)$ lies on it.

Show Guidance/Answer

Equation: All points $(x,y)$ satisfying

${(x-3)}^2+{(y--2)}^2=5^2$ $\Rightarrow$ ${(x-3)}^2+{(y+2)}^2=25$

Check the point $(7,-2)$:

${(7-3)}^2+{(-2+2)}^2=4^2+0^2=16$

Since $16$ is not $25$, the point is inside the circle but not on it.

Extension 3 – Classify a triangle (acute, right, obtuse)

Use the distance formula to classify triangle $PQR$ with $P=(0,0)$, $Q=(6,3)$, $R=(2,9)$. Hint: compare the largest squared side with the sum of the other two squared sides.

Show Guidance/Answer

Compute squared lengths:

${\text{PQ}}^2=6^2+3^2=45$, ${\text{PR}}^2=2^2+9^2=85$, ${\text{QR}}^2={(-4)}^2+6^2=52$.

Compare: largest is ${\text{PR}}^2=85$. Sum of the other two is $45+52=97$.

Since $85$ is less than $97$, the triangle is acute (all angles less than 90°).

Extension 4 – Find the fourth vertex of a rectangle

Three vertices of a rectangle are $A=(1,2)$, $B=(7,2)$, and $C=(7,8)$. Find the fourth vertex $D$ and verify the rectangle using distances.

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Method: Opposite sides are parallel and equal; use the parallelogram rule $D=A+C-B$ (component-wise).

Compute D: $D=(1+7-7,2+8-2)=(1,8)$.

Verify diagonals equal:

$\text{AC}:\sqrt{{(7-1)}^2+{(8-2)}^2}=\sqrt{36+36}=6\sqrt{2}$

$\text{BD}:\sqrt{{(1-7)}^2+{(8-2)}^2}=\sqrt{36+36}=6\sqrt{2}$

Diagonals equal $\Rightarrow$ rectangle confirmed.

Extension 5 – Perpendicular bisector via equal distances

Points $A=(-2,3)$ and $B=(6,3)$. Find the equation of the set of all points equidistant from A and B (the perpendicular bisector of segment AB). Then check that the midpoint lies on your line.

Show Guidance/Answer

Equal-distance condition: For any $(x,y)$, set

$\sqrt{{(x+2)}^2+{(y-3)}^2}=\sqrt{{(x-6)}^2+{(y-3)}^2}$

Simplify: Square both sides and cancel the identical ${(y-3)}^2$ terms:

${(x+2)}^2={(x-6)}^2\Rightarrow x+2=\pm (x-6)$

The + case gives $2=-6$ (false). The − case gives

$x+2=-x+6$ $\Rightarrow$ $2x=4$ $\Rightarrow$ $x=2$.

Therefore the perpendicular bisector is the vertical line $x=2$.

Midpoint check: Midpoint of AB is $(\frac{-2+6}{2},\frac{3+3}{2})=(2,3)$, which satisfies $x=2$.

Write your answers in your notebook. Use complete sentences and include quick sketches where helpful.

1. In your own words, explain how the distance formula comes from the Pythagorean Theorem. Draw and label a right triangle formed by two points on a grid. Clearly mark the horizontal and vertical changes and the hypotenuse.
2. Pick two real places you know on campus or in your neighborhood. Make a simple coordinate grid for your sketch, assign reasonable coordinates, and choose a scale. Compute the straight-line distance using the distance formula. State the exact form and a rounded decimal with units. Briefly explain how your choice of scale and rounding affects the result.
3. Describe a situation where city-walk distance and straight-line distance are different. Create a small numerical example to show the two values and explain which is more appropriate and why.
4. Identify one mistake you nearly made today (for example, adding before squaring, sign errors, or rounding too early). Explain how you could catch that mistake quickly in the future.
5. Self-assess the Day 2 objectives:
   • I can find a missing side or distance between points using the distance formula.
   • I can derive or justify the distance formula from the Pythagorean Theorem.
   • I can interpret answers with correct units, scale, and reasonable rounding.
   For each, rate yourself as Needs practice, Getting there, or Confident. Write one concrete action you will take before the next lesson.