Step 3
Subtract Inside
Watch double negatives!
Step 4
Square Both
Result is ALWAYS positive
Step 5
Add Under \(\sqrt{\ }\)
Leave in radical form
I DO: Fully Worked Example Quiz Problem #1 Direct Model
Problem: Calculate the length of segment \(AB\) with endpoints A(-10, 4) and B(11, 9).
Steps 1 & 2: Label & Setup
\(x_1 = \mathbf{-10}\), \(y_1 = \mathbf{4}\)
\(x_2 = \mathbf{11}\), \(y_2 = \mathbf{9}\)
\(d = \sqrt{(11 - (-10))^2 + (9 - 4)^2}\)
Step 3: Subtract Inside
\(11 - (-10) = 11 + 10 = \mathbf{21}\)
\(9 - 4 = \mathbf{5}\)
\(d = \sqrt{(21)^2 + (5)^2}\)
Step 4: Square Each
\(21^2 = 21 \times 21 = \mathbf{441}\)
\(5^2 = 5 \times 5 = \mathbf{25}\)
\(d = \sqrt{441 + 25}\)
Step 5: Add
\(d = \sqrt{466}\)
Choice (C) on Quiz!
WE DO: Scaffolded Practice ANSWER KEY
Solved: \(d = 13\)
Problem: Calculate the length of segment \(CD\) with endpoints C(-4, 2) and D(8, 7).
1. Label Points:
\(x_1 =\) -4, \(y_1 =\) 2
\(x_2 =\) 8, \(y_2 =\) 7
2. Substitute:
\(d = \sqrt{(\mathbf{8} - (\mathbf{-4}))^2 + (\mathbf{7} - \mathbf{2})^2}\)
3 & 4. Subtract & Square:
\(8 - (-4) =\) 12 → sq: 144
\(7 - 2 =\) 5 → sq: 25
5. Final Answer:
\(d = \sqrt{\mathbf{169}} = \mathbf{13}\)
YOU DO: Independent Application ANSWER KEY
Solution: \(d = 13\)
Calculate the distance between P(-6, 3) and Q(6, 8). Follow the 5 ladder steps above.
Complete Worked Steps:
1. \(x_1 = -6, y_1 = 3\) and \(x_2 = 6, y_2 = 8\)
2. \(d = \sqrt{(6 - (-6))^2 + (8 - 3)^2} = \sqrt{(12)^2 + (5)^2}\)
3. \(d = \sqrt{144 + 25} = \sqrt{169} = \mathbf{13}\)
Exact Distance Result:
\(d = \sqrt{\mathbf{169}} = \mathbf{13}\)
Geometry Foundations Packet • Teacher Answer Key Page 2 of 5 • Skill 3 Solutions
Teacher Answer Key Coordinate Algebra
Teacher Master Key • Full Algebraic Steps • Red Ink Annotations
Part 3 of 4
Key • Page 3
Core Geometric Law
Parallel Lines (\(\parallel\)) have the EXACT SAME SLOPE!
\(m_1 = m_2\)
Never change the sign or flip parallel slope!
3-Step Action Plan
1. Solve for \(y\)
Get \(y = \mathbf{m}x + b\) to uncover the slope \(\mathbf{m}\).
2. Point-Slope Form
Plug point \((x_1, y_1)\) and slope \(\mathbf{m}\) into:
\(y - y_1 = m(x - x_1)\)
3. Distribute & Solve
Distribute \(\mathbf{m}\), then add \(y_1\) to isolate \(y\).
I DO: Fully Worked Example Quiz Problem #3 Direct Model
Write the equation of the line that is parallel to 2y + 6 = 10x and passes through the point (4, 7).
Step 1: Find the Slope \(m\)
\(2y + 6 = 10x\)
\(2y = 10x - 6\)
\(y = \frac{10}{2}x - \frac{6}{2}\)
\(y = \mathbf{5}x - 3 \implies m = \mathbf{5}\)
Step 2: Point-Slope Formula
Point: \((x_1, y_1) = (4, 7)\)
Slope: \(m = 5\)
\(y - y_1 = m(x - x_1)\)
\(y - 7 = 5(x - 4)\)
Step 3: Solve for \(y\)
\(y - 7 = 5x - 20\)
\(\phantom{y - } +7 \quad\quad\ \ +7\)
\(y = 5x - 13\)
Matches Choice (C)!
WE DO: Guided Practice ANSWER KEY
Answer: \(y = 2x + 4\)
Write the equation of the line parallel to 3y - 9 = 6x passing through (2, 8).
1. Uncover Slope \(m\):
\(3y = 6x +\) 9
\(y =\) 2\(x + 3\)
Slope \(m =\) 2
2. Plug into Point-Slope:
\(y - y_1 = m(x - x_1)\)
\(y -\) 8 \(= 2(x -\) 2\()\)
\(y - 8 = 2x -\) 4
3. Add 8 to both sides:
Final Equation:
\(y = 2x +\) 4
YOU DO: Independent Application ANSWER KEY
Solution: \(y = 3x - 4\)
Write the equation of the line parallel to 2y + 8 = 6x and passes through (3, 5).
Full Worked Steps:
1. \(2y = 6x - 8 \implies y = 3x - 4 \implies \mathbf{m = 3}\)
2. \(y - 5 = 3(x - 3) \implies y - 5 = 3x - 9\)
3. \(y = 3x - 9 + 5 \implies \mathbf{y = 3x - 4}\)
Parallel Line Equation:
\(y = \mathbf{3x - 4}\)
Geometry Foundations Packet • Teacher Answer Key Page 3 of 5 • Skill 4 Solutions
Teacher Answer Key Algebra & Geometry Modeling
Teacher Master Key • Factoring & Dimensions Solutions • Red Ink Annotations
Part 4 of 4
Key • Page 4
Quiz #5 Direct Model
I DO: \(y = 6x^2 + x - 2\)
1. Multiply \(a \cdot c\): \(6 \cdot (-2) = \mathbf{-12}\)
2. Find 2 numbers that multiply to \(-12\) and add to \(b = 1\): \(+4\) and \(-3\)
3. Split & Factor by grouping:
\(6x^2 + 4x - 3x - 2 = 0\)
\(2x(3x + 2) - 1(3x + 2) = 0\)
\((2x - 1)(3x + 2) = 0\)
4. Solve each factor for zero:
\(2x - 1 = 0 \implies x = \mathbf{\frac{1}{2}}\)
\(3x + 2 = 0 \implies x = \mathbf{-\frac{2}{3}}\)
Roots: \((\frac{1}{2}, 0)\) and \((-\frac{2}{3}, 0)\) (Choice B)
WE DO: \(y = 2x^2 + 5x + 2\) ANSWER KEY
1. Product \(a \cdot c = 2 \cdot 2 = 4\). Sum \(= 5\).
Two numbers: 4 and 1
2. Factored form: \((2x + 1)(x + \mathbf{2}) = 0\)
3. Solve each for \(x\):
\(2x + 1 = 0 \implies x =\) -\(\frac{1}{2}\)
\(x + 2 = 0 \implies x =\) -2
Roots Written as Coordinates:
\((\mathbf{-\frac{1}{2}},\ 0)\) and \((\mathbf{-2},\ 0)\)
Quiz #2 Direct Model
Quiz Problem #2: The length of a rectangle is \(x\) ft, the width is \((x + 4)\) ft, and the area is \(32\text{ ft}^2\). Determine the measure of the WIDTH.
Step 1: Formula Setup
\(\text{Area} = \text{length} \times \text{width}\)
\(x(x + 4) = 32\)
\(x^2 + 4x - 32 = 0\)
Step 2: Factor Equation
Multiply to \(-32\), add to \(4\):
Factors: \(+8\) and \(-4\)
\((x + 8)(x - 4) = 0\)
Step 3: Solve for \(x\)
Length cannot be negative!
\(x = -8\) (discard)
\(x = 4\text{ ft (Length)}\)
Step 4: Answer the Question!
Question asks for WIDTH:
\(\text{Width} = x + 4\)
\(\text{Width} = 4 + 4 = \mathbf{8}\text{ ft}\)
Careful trap: don't stop at x = 4!
Your Turn: A rectangle has length \(x\) and width \((x + 2)\). Area \(= 24\text{ ft}^2\).
ANSWER KEY
1. \(x(x + 2) = 24 \implies x^2 + 2x - 24 = 0\)
2. \((x + 6)(x - 4) = 0 \implies x =\) 4
Width \(= x + 2 =\) 6 ft
Geometry Foundations Packet • Teacher Answer Key Page 4 of 5 • Skills 5 & 6 Solutions
Teacher Answer Key Exit Ticket Master
Correct Options Highlighted • Step-by-Step Work Filled Directly in Templates
Scored Key 5 / 5 (100%)
Name: TEACHER MASTER KEY
Date: Exit Assessment
Period: All Periods
1 Which of the following is a proper name for the ray shown below?
R S T
Key Note: Solid endpoint is R, so ray must begin with letter R: \(\vec{RS}\) or \(\vec{RT}\).
(A) \(\vec{RS}\)
(B) \(\vec{SR}\)
(C) \(\vec{TR}\)
(D) \(\overline{RT}\)
2 In the diagram, which set represents collinear points?
P Q R S W Z
(A) W, Q, R
(B) P, Q, S
(C) P, Z, S
(D) W, R, Z
3 Calculate the length of segment \(AB\) with endpoints A(-5, 2) and B(7, 7).
Formula Solution Steps:
\(d = \sqrt{(7 - (-5))^2 + (7 - 2)^2} = \sqrt{(12)^2 + (5)^2}\)
\(d = \sqrt{144 + 25} = \mathbf{\sqrt{169} = 13}\)
(A) \(\sqrt{119}\)
(B) \(\sqrt{169}\)
(C) \(\sqrt{193}\)
(D) \(\sqrt{205}\)
4 Which equation represents the line that is parallel to 2y + 4 = 8x and passes through the point (2, 3)?
Slope & Point-Slope Steps:
\(2y = 8x - 4 \implies y = 4x - 2 \implies \mathbf{m = 4}\)
\(y - 3 = 4(x - 2) \implies y - 3 = 4x - 8 \implies \mathbf{y = 4x - 5}\)
(A) \(y = 4x - 5\)
(B) \(y = 4x + 3\)
(C) \(y = 8x - 13\)
(D) \(y = 8x - 5\)
5 The length of a rectangle is \(x\) ft, the width is \((x + 3)\) ft, and the area is \(40\text{ ft}^2\). What is the measure of the WIDTH?
Factoring & Dimension Steps:
\(x(x + 3) = 40 \implies x^2 + 3x - 40 = 0 \implies (x + 8)(x - 5) = 0 \implies x = 5\)
\(\text{Width} = x + 3 = 5 + 3 = \mathbf{8\text{ ft}}\) (Length is 5 ft)
(A) 5 ft
(B) 8 ft
(C) 10 ft
(D) 13 ft
Mastery Benchmark: 4 or 5 correct indicates readiness for Unit 1 Exam without modifications.
1. A 2. B 3. B 4. A 5. B
