Honors Geometry Unit 6 Review: Coordinate Geometry Lessons 1-15

Unit 6 Coordinate Geometry Unit Review

Name: Alyson Ankrab
Unit Content: Lessons 1 through 15

Lessons 1-3: Coordinate Transformation Notation

  • Objective: Use coordinate transformation notation to take points in the plane as inputs and provide other points as outputs.
Transformation Example 1
  • Transformation Rule: (x,y)→(x+4,−y)(x, y) \rightarrow (x + 4, -y)
  • Description of Transformation:     * Translation: 4 units to the right.     * Reflection: Reflected across the x-axis.
  • Coordinate Changes for Image Processing:     * Original Point 1: (3,2)(3, 2), Image Point: (7,−2)(7, -2)     * Original Point 2: (−4,1)(-4, 1), Image Point: (0,−1)(0, -1)     * Original Point 3: (0,3)(0, 3), Image Point: (4,−3)(4, -3)
Classification of Transformations
  • Objective: Determine whether a transformation produces a congruent, similar, or neither (non-congruent/non-similar) image.
  • Transformations producing Congruent Images:     * Translations     * Reflections     * Rotations
  • Transformations producing Similar Images:     * Dilations
  • Transformations producing Neither:     * Stretches     * Shrinks
Coordinate Rule Analysis

For each rule below, the resulting image is classified as Congruent (C), Similar (S), or Neither (N):

  • (x,y)→(2x,y)(x, y) \rightarrow (2x, y): Neither (N)
  • (x,y)→(y,2x)(x, y) \rightarrow (y, 2x): Neither (N)
  • (x,y)→(y,−x)(x, y) \rightarrow (y, -x): Congruent (C)
  • (x,y)→(−x,y)(x, y) \rightarrow (-x, y): Congruent (C)
  • (x,y)→(x2,y2)(x, y) \rightarrow (\frac{x}{2}, \frac{y}{2}): Similar (S)
  • (x,y)→(−x,−y)(x, y) \rightarrow (-x, -y): Congruent (C)
  • (x,y)→(−2x,y)(x, y) \rightarrow (-2x, y): Neither (N)

Lesson 4: Equation of a Circle

  • Objective: Write the equation of a circle given its center and radius.
  • Standard Form Equation of a Circle:(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2     * Center (h,k)(h, k)     * Radius rr
Circle Construction Example
  • Given: Center at (−4,3)(-4, 3) and a radius of 55.
  • Draft Equation: (x−(−4))2+(y−3)2=52(x - (-4))^2 + (y - 3)^2 = 5^2
  • Final Form: (x+4)2+(y−3)2=25(x + 4)^2 + (y - 3)^2 = 25

Lessons 5-6: Completing the Square for Circles

  • Objective: Complete the square to find the standard form of a circle equation and determine the center and radius.
Example A
  • Equation: x2+6x+9+y2−8y+16=36x^2 + 6x + 9 + y^2 - 8y + 16 = 36
  • Factored Form: (x+3)2+(y−4)2=36(x + 3)^2 + (y - 4)^2 = 36
  • Resulting Center: (−3,4)(-3, 4)
  • Resulting Radius: 36=6\sqrt{36} = 6
Example B
  • Equation: x2−10x+y2+4y=7x^2 - 10x + y^2 + 4y = 7
  • Transformation Steps:     * Grouping: (x2−10x+25)+(y2+4y+4)=7+25+4(x^2 - 10x + 25) + (y^2 + 4y + 4) = 7 + 25 + 4     * (x−5)2+(y+2)2=36(x - 5)^2 + (y + 2)^2 = 36
  • Resulting Center: (5,−2)(5, -2)
Example C
  • Equation Segment: x2+y2−10x−2y−10=0x^2 + y^2 - 10x - 2y - 10 = 0
  • Resulting Calculation: The coordinates noted indicate r=0r = 0 in the provided worksheet scribbles, though the full expansion is obscured.
Example D
  • Equation: x2+y2−6x−8y−8=0x^2 + y^2 - 6x - 8y - 8 = 0
  • Center/Radius: Annotated as r=0r = 0 based on partial manual work.

Lesson 9: Point-Slope Form of a Line

  • Objective: Use the definition of slope to write the equation of a line in point-slope form or slope-intercept form.
  • Point-Slope Form Formula:y−y1=m(x−x1)y - y_1 = m(x - x_1)     * mm is the slope of the line.     * (x1,y1)(x_1, y_1) is a point the line passes through.
Linear Equation Examples
  1. Slope calculation: Line with slope of 22 passing through (−3,4)(-3, 4).     * Equation: y−4=2(x−(−3))y - 4 = 2(x - (-3))
  2. Origin/Point calculation: Line with slope of −2-2 passing through (0,5)(0, 5).     * Equation: y−5=−2(x−0)y - 5 = -2(x - 0)

Lesson 10: Parallel Lines

  • Objectives: Find the slope of a line parallel to another and write equations of parallel lines given an equation and a point.
Parallel Line Exercises
  1. Requirement: Line parallel to y=4x−3y = 4x - 3 passing through (−2,1)(-2, 1).     * The slope of the new line is 44.
  2. Requirement: Line parallel to y=2x−3y = 2x - 3 with a y-intercept of 66.     * The point is (0,6)(0, 6) and the slope is 22.

Lesson 11: Perpendicular Lines

  • Objectives: Find the slope of a line perpendicular to another and write equations given a point.
Perpendicular Line Exercises
  1. Requirement: Line perpendicular to y=x+5y = x + 5 passing through (4,−1)(4, -1).
  2. Requirement: Line perpendicular to y=x+3y = x + 3 passing through the origin (0,0)(0, 0).
  3. Requirement: Line perpendicular to y=x+7y = x + 7 passing through (4,−3)(4, -3).
Analysis of Line Relationships

Line m Equation: y−2=32(x+3)y - 2 = \frac{3}{2}(x + 3) Identify the relationship (Parallel, Perpendicular, or Neither):

  • A. y=32x−6y = \frac{3}{2}x - 6: Parallel
  • B. y=32x+1y = \frac{3}{2}x + 1: Parallel
  • C. y=−23x−3y = -\frac{2}{3}x - 3: Perpendicular
  • D. y=32x+8y = \frac{3}{2}x + 8: Parallel
  • E. y−3=−23(x−2)y - 3 = -\frac{2}{3}(x - 2): Perpendicular
  • F. y+1=32(x+4)y + 1 = \frac{3}{2}(x + 4): Parallel

Lesson 13: Intersection Points

  • Objective: Find and verify the intersection points of a line and a circle.
  • System Setup:     * Circle Equation: (x+2)2+(y+4)2=16(x + 2)^2 + (y + 4)^2 = 16     * Line Equation: y=32x−3y = \frac{3}{2}x - 3
Verification Work
  1. Is (2,−4)(2, -4) an intersection point?     * Verification for the circle: (2+2)2+(−4+4)2=42+02=16(2 + 2)^2 + (-4 + 4)^2 = 4^2 + 0^2 = 16. (True)     * Verification for the line: −4=32(2)−3=3−3=0-4 = \frac{3}{2}(2) - 3 = 3 - 3 = 0. (False: −4≠0-4 \neq 0)     * Result: It is NOT an intersection point.
  2. Is (−6,0)(-6, 0) an intersection point? (Evaluation pending from graph).

Lesson 14: Geometric Theorems with Coordinates

  • Objective: Use coordinates of figures to prove geometric theorems.
  • Proof Exercise: Prove quadrilateral HYPE is a rectangle.     * Vertices: H(−3,6)H(-3, 6), Y(2,9)Y(2, 9), P(8,−1)P(8, -1), and E(3,−4)E(3, -4).     * Method: Establish that opposite sides have the same slope and that adjacent sides have negative reciprocal slopes (forming right angles).

Lesson 15: Partitioning Line Segments

  • Objective: Calculate coordinates of a point PP on a line segment that partitions it in a given ratio.
Partitioning Example 1
  • Segment: Directed line segment from MM to EE.     * Initial Point M: (7,−2)(7, -2)     * Terminal Point E: (−5,6)(-5, 6)
  • Ratio: 1:31:3
  • Formula Calculation Scribed: 34E+14M\frac{3}{4}E + \frac{1}{4}M     * 34(−5,6)+14(7,−2)=(−3.75,4.5)+(1.75,−0.5)=(−2,4)\frac{3}{4}(-5, 6) + \frac{1}{4}(7, -2) = (-3.75, 4.5) + (1.75, -0.5) = (-2, 4)
  • Partition Point P Result: (−2,5)(-2, 5) (as annotated in review sheet).
Partitioning Example 2
  • Segment: Directed line segment from AA to BB.     * Initial Point A: (−8,−5)(-8, -5)     * Terminal Point B: (8,3)(8, 3)
  • Ratio: 5:35:3
  • Formula Calculation Scribed: 38A+58B\frac{3}{8}A + \frac{5}{8}B
  • Partition Point P Result: (2,1.25)(2, 1.25), or roughly (2,0)(2, 0) depending on rounding in the scribe work.