Geometry Foundations: Segment & Angle Measurement, Basic Constructions, Midpoint & Distance

Lesson 1-1: Measuring Segments and Angles

  • Learning Goals and Standards

    • Learning Objective: Use properties of segments and angles to find their measures.

    • Texas Essential Knowledge and Skills (TEKS):

    • G.4AG.4A: Distinguish between undefined terms, definitions, postulates, conjectures, and theorems.

    • G.2AG.2A: Determine coordinates of a point that is a given fractional distance less than one from one end of a line segment to the other.

    • Mathematical Process Standards: G.1AG.1A, G.1BG.1B, G.1CG.1C, G.1DG.1D, G.1FG.1F, G.1GG.1G.

    • Essential Question: How are the properties of segments and angles used to determine their measures?

    • Core Vocabulary: collinear points, line, plane, point, postulate.

  • Undefined Terms

    • Undefined terms are terms whose meanings are accepted without formal definition. They serve as the foundational building blocks of geometry.

    • Point:

    • Description: A point is a location and has no size.

    • Diagram: Represented by a dot labelled with point PP.

    • Notation: A single capital letter, such as PP.

    • Line:

    • Description: A line consists of infinitely many points on a straight path that extends in two opposite directions with no end and no thickness.

    • Diagram: A straight path with arrowheads on both ends containing points AA and BB, or designated by line ℓ\ell.

    • Notation: A single lowercase letter like line ℓ\ell, or any two points on the line written beneath a double-sided arrow symbol: AB↔\overleftrightarrow{AB}.

    • Plane:

    • Description: A plane consists of infinitely many points and lines on a flat surface that extends without end and has no thickness.

    • Diagram: A slanted four-sided flat surface designated as plane MM or containing non-collinear points XX, YY, and Z$.\n - Notation: A single capital letter such as plane M,orthreepointsintheplanethatdonotlieonthesameline:plane, or three points in the plane that do not lie on the same line: planeXYZ$.

  • Defined Terms

    • Defined terms are terms constructed using previously defined or known geometric terms.

    • Segment (Line Segment):

    • Description: Part of a line consisting of two endpoints and all points between them.

    • Notation: Named by its two endpoints with a solid bar over them, such as AB‾\overline{AB}. Length is written without the bar: AB$.\n - Ray:\n - Description: Part of a line consisting of one endpoint and all the points of the line on one side of the endpoint.\n - Notation: Named using its endpoint first followed by any other point on the ray with a single-direction arrow: \overrightarrow{MN}.\n - Opposite Rays:\n - Description: Rays with the exact same endpoint that lie on the same line extending in opposite directions.\n - Example: Given points S,,T,and, andUonalineinorder,theoppositeraysareon a line in order, the opposite rays are\overrightarrow{TS}andand\overrightarrow{TU}.\n - Angle:\n - Description: Formed by two rays with the same endpoint. Each ray is a side of the angle, and the common endpoint is the vertex of the angle.\n - Diagram: Vertex Qconnectedtosideraysconnected to side rays\overrightarrow{QP}andand\overrightarrow{QR},withaninteriorlabel, with an interior label2$.

    • Notation: Named by its vertex ∠Q\angle Q, three points with the vertex listed in the middle ∠PQR\angle PQR, or an interior number \angle 2$.\n\n- Measuring Segment Lengths\n - Definition of Segment Length: The length of a segment is a positive real number representing the distance between its endpoints.\n - Postulate 1-1 (Ruler Postulate):\n - Every point on a line can be paired with a unique real number called the coordinate of the point.\n - The distance between two points is the absolute value of the difference between their coordinates.\n - Example: Point Xhascoordinatehas coordinate3\,cmandPointand PointYhascoordinatehas coordinate7\,cm.Thedistanceis. The distance isXY = |7 - 3| = 4\,cm.\n - Example 1 Calculations:\n - On a number line with points A(-3),,B(-1),,C(1),and, andD(4):\n - Length CD = |4 - 1| = 3\n - Length AC = |1 - (-3)| = |1 + 3| = 4\n - Example 2 Calculations:\n - Coordinates on a centimeter scale: K = 12,,L = 16,,M = 27.\n - KM = |27 - 12| = 15ororKM = |12 - 27| = 15\n - LM = |27 - 16| = 11ororLM = |16 - 27| = 11\n - KL = |16 - 12| = 4ororKL = |12 - 16| = 4\n - Key Principle: Distance is strictly positive; always evaluate distance as the absolute value of the coordinate difference.\n - Notation Distinction:\n - \overline{AB}referstothegeometricobject(segmentrefers to the geometric object (segmentAB).\n - ABreferstothenumericalvalue(themeasureorlengthofrefers to the numerical value (the measure or length of\overline{AB}).\n\n- Segment Addition\n - Collinear Points: Points that lie on the same line.\n - Postulate 1-2 (Segment Addition Postulate):\n - If points A,,B,and, andCareonthesamelinewithare on the same line withBbetweenbetweenAandandC,then, thenAB + BC = AC$.

    • Example 3 (Solving Algebraic Segments):

    • Given collinear points FF, GG, and HH with FG=3x−1FG = 3x - 1, GH=2x+2GH = 2x + 2, and GH=16GH = 16.

    • Step 1: Solve for xx using GHGH:       2x+2=16  ⟹  2x=14  ⟹  x=72x + 2 = 16 \implies 2x = 14 \implies x = 7

    • Step 2: Calculate total length FHFH using the Segment Addition Postulate:       FH=FG+GH=(3x−1)+(2x+2)=5x+1FH = FG + GH = (3x - 1) + (2x + 2) = 5x + 1       FH=5(7)+1=35+1=36FH = 5(7) + 1 = 35 + 1 = 36

    • Common Error Warning: Do not mistake the value of xx for the final answer when the problem asks for a segment length.

    • Try It 3 Questions:

    • Points JJ, KK, and LL are collinear with JK=3nJK = 3n, KL=5n−7KL = 5n - 7, and JL = 25$.\n - Part a: Find n:\n      3n + (5n - 7) = 25 \implies 8n - 7 = 25 \implies 8n = 32 \implies n = 4\n - Part b: Find JKandandKL:\n      JK = 3(4) = 12\n      KL = 5(4) - 7 = 20 - 7 = 13\n\n- Angle Measurement and Protractor Use\n - Protractor Postulate (Example 4):\n - Real numbers are assigned to rays extending from a central vertex.\n - Given ray \overrightarrow{EA}alignedwithaligned with0^\circ on the protractor scale:\n - Ray \overrightarrow{EB}isassignedis assigned47^\circ.\n - Ray \overrightarrow{EC}isassignedis assigned105^\circ.\n - The measure of \angle BEC((m\angle BEC) is evaluated as the absolute difference between ray values:\n      m\angle BEC = |105^\circ - 47^\circ| = 58^\circ\n      Alternatively: m\angle BEC = m\angle AEC - m\angle AEB = 105^\circ - 47^\circ = 58^\circ\n - Postulate 1-4 (Angle Addition Postulate):\n - If point Disintheinteriorofis in the interior of\angle ABC,then, thenm\angle ABD + m\angle DBC = m\angle ABC$.

  • Lesson 1-1 Practice Exercises and Solutions

    • Number Line Calculations (W=−4W = -4, X=−1X = -1, Y=4Y = 4):

    • Segment WX=∣−1−(−4)∣=∣−1+4∣=3WX = |-1 - (-4)| = |-1 + 4| = 3

    • Segment WY=∣4−(−4)∣=∣4+4∣=8WY = |4 - (-4)| = |4 + 4| = 8

    • Collinear Points A,B,CA, B, C with BB between AA and CC:

    • Given AB=12AB = 12 and AC=19AC = 19:       BC=AC−AB=19−12=7BC = AC - AB = 19 - 12 = 7

    • Angle Addition Calculations:

    • Given m∠JML=80∘m\angle JML = 80^\circ and m∠KML=33∘m\angle KML = 33^\circ:       m∠JMK=m∠JML−m∠KML=80∘−33∘=47∘m\angle JMK = m\angle JML - m\angle KML = 80^\circ - 33^\circ = 47^\circ

    • Additional Practice Set:

    • Number Line Points (R=−5R = -5, S=−2S = -2, T=2T = 2, U=6U = 6):

      • RS=∣−2−(−5)∣=3RS = |-2 - (-5)| = 3

      • RT=∣2−(−5)∣=7RT = |2 - (-5)| = 7

      • ST=∣2−(−2)∣=4ST = |2 - (-2)| = 4

      • RU=∣6−(−5)∣=11RU = |6 - (-5)| = 11

    • Collinear Points P,Q,RP, Q, R with PQ=3x−5PQ = 3x - 5, QR=14QR = 14, PR=5x−1PR = 5x - 1:       (3x−5)+14=5x−1  ⟹  3x+9=5x−1  ⟹  2x=10  ⟹  x=5(3x - 5) + 14 = 5x - 1 \implies 3x + 9 = 5x - 1 \implies 2x = 10 \implies x = 5

      • PQ=3(5)−5=10PQ = 3(5) - 5 = 10

      • QR=14QR = 14

      • PR=5(5)−1=24PR = 5(5) - 1 = 24

    • Collinear Points A,B,C,DA, B, C, D:

      • If AC=24AC = 24, AB=3xAB = 3x, and BC=4x−13BC = 4x - 13:         3x+(4x−13)=24  ⟹  7x−13=24  ⟹  7x=37  ⟹  x=3773x + (4x - 13) = 24 \implies 7x - 13 = 24 \implies 7x = 37 \implies x = \frac{37}{7}         AB=3(377)=1117AB = 3\left(\frac{37}{7}\right) = \frac{111}{7}

      • If BC=15BC = 15:         4x−13=15  ⟹  4x=28  ⟹  x=74x - 13 = 15 \implies 4x = 28 \implies x = 7         If CD=2x+1CD = 2x + 1, then BD=BC+CD=15+(2(7)+1)=15+15=30BD = BC + CD = 15 + (2(7) + 1) = 15 + 15 = 30

    • Error Analysis (Luis Angle Calculation):

      • Luis incorrectly stated that m∠QTR=80∘m\angle QTR = 80^\circ. His error was reading the incorrect scale on the protractor (mixing the inner scale reading with the outer scale reading rather than taking the absolute difference along a single consistent scale).

    • Newscast Camera Application:

      • Three cameras setup: Center camera faces anchor desk. Two side cameras angled 45∘45^\circ away from center camera. Each camera has a 60∘60^\circ field of view.

      • The outer boundary of each side camera extends 45∘+30∘=75∘45^\circ + 30^\circ = 75^\circ to the left and right of center.

      • Total span covered by all three cameras = 75∘+75∘=150∘75^\circ + 75^\circ = 150^\circ.

Lesson 1-2: Basic Constructions

  • Objectives and Standards

    • Learning Objective: Use a straightedge and compass to construct basic figures.

    • Core Vocabulary: angle bisector, construction.

    • TEKS G.5BG.5B: Construct congruent segments, congruent angles, a segment bisector, an angle bisector, perpendicular lines, the perpendicular bisector of a line segment, and a line parallel to a given line through a point not on a line using a compass and a straightedge.

    • Mathematical Process Standards: G.1CG.1C, G.1DG.1D, G.1EG.1E, G.1FG.1F, G.1GG.1G.

    • Essential Question: How are a straightedge and compass used to make basic constructions?

  • Geometric Congruence

    • Congruent Segments: Segments that have identical length.

    • Statement: AB‾≅CD‾  ⟺  AB=CD\overline{AB} \cong \overline{CD} \iff AB = CD

    • Visual Representation: Indicated by matching tick marks across segments.

    • Congruent Angles: Angles that have identical measure.

    • Statement: ∠TUV≅∠XYZ  ⟺  m∠TUV=m∠XYZ\angle TUV \cong \angle XYZ \iff m\angle TUV = m\angle XYZ

    • Visual Representation: Indicated by matching arc marks within angles.

    • Properties of Congruence:

    • Reflexive Property of Congruence:       AB‾≅AB‾\overline{AB} \cong \overline{AB}       ∠A≅∠A\angle A \cong \angle A

    • Symmetric Property of Congruence:       If AB‾≅CD‾, then CD‾≅AB‾\text{If } \overline{AB} \cong \overline{CD}\text{, then } \overline{CD} \cong \overline{AB}       If ∠A≅∠B, then ∠B≅∠A\text{If } \angle A \cong \angle B\text{, then } \angle B \cong \angle A

    • Transitive Property of Congruence:       If AB‾≅CD‾ and CD‾≅EF‾, then AB‾≅EF‾\text{If } \overline{AB} \cong \overline{CD} \text{ and } \overline{CD} \cong \overline{EF}\text{, then } \overline{AB} \cong \overline{EF}       If ∠A≅∠B and ∠B≅∠C, then ∠A≅∠C\text{If } \angle A \cong \angle B \text{ and } \angle B \cong \angle C\text{, then } \angle A \cong \angle C

  • Congruence Examples

    • Example 1 Part A:

    • Given m∠XWZ=127∘m\angle XWZ = 127^\circ, m∠XWY=32∘m\angle XWY = 32^\circ, and m∠VWZ=32∘m\angle VWZ = 32^\circ. Find m∠YWVm\angle YWV.

    • Apply Angle Addition Postulate:       m∠XWY+m∠YWV+m∠VWZ=m∠XWZm\angle XWY + m\angle YWV + m\angle VWZ = m\angle XWZ       32∘+m∠YWV+32∘=127∘32^\circ + m\angle YWV + 32^\circ = 127^\circ       64∘+m∠YWV=127∘  ⟹  m∠YWV=63∘64^\circ + m\angle YWV = 127^\circ \implies m\angle YWV = 63^\circ

    • Example 1 Part B:

    • Find length HFHF given tick mark congruences: HG‾≅AH‾\overline{HG} \cong \overline{AH} (AH=11 cmAH = 11\,cm) and GF‾≅BC‾\overline{GF} \cong \overline{BC} (BC=8 cmBC = 8\,cm).

    • Apply Segment Addition Postulate:       HF=HG+GF=AH+BC=11 cm+8 cm=19 cmHF = HG + GF = AH + BC = 11\,cm + 8\,cm = 19\,cm

    • Try It 1 Questions:

    • Part a: Given m∠NOP=2x+2m\angle NOP = 2x + 2, m∠POR=3x−5m\angle POR = 3x - 5, and m∠NOQ=114∘m\angle NOQ = 114^\circ. If ∠NOP≅∠QOR\angle NOP \cong \angle QOR, solve for xx using angle relationships.

    • Part b: Given CD=11.5 cmCD = 11.5\,cm, DE=5.3 cmDE = 5.3\,cm, and perimeter =73.8 cm= 73.8\,cm, evaluate length GEGE.

  • Construction Rules and Definitions

    • Construction: A geometric figure drawn using strictly an unmarked straightedge and a compass.

    • Straightedge: A tool used solely to draw straight lines, rays, or line segments (not for measuring lengths).

    • Compass: A tool used to draw arcs and circles of defined radii, and to copy segment lengths.

  • Step-by-Step Construction Procedures

    • Copying a Line Segment (Copying AB‾\overline{AB} to form MN‾\overline{MN}):

    • Step 1: Use a straightedge to draw line ℓ\ell. Mark point MM on line \ell$.\n - Step 2: Place the compass point at A,andadjusttheopeningtolength, and adjust the opening to lengthAB$.

    • Step 3: Keeping the exact compass setting, place the point at MM and draw an arc intersecting line ℓ\ell. Label the intersection point N$.\n - Result: Constructed segment \overline{MN} \cong \overline{AB}.\n - Copying an Angle (Copying \angle Atoformto form\angle YXZ):\n - Step 1: Mark point X.Useastraightedgetodrawarayextendingfrom. Use a straightedge to draw a ray extending fromX$.

    • Step 2: Place the compass point at AA. Draw an arc intersecting both rays of ∠A\angle A. Label intersection points BB and C$.\n - Step 3: Without changing the compass setting, place the point at Xanddrawanarcintersectingthenewray.Labelintersectionpointand draw an arc intersecting the new ray. Label intersection pointY$.

    • Step 4: Place compass point at CC, and open the compass setting to distance BC$.\n - Step 5: Without changing the setting, place the compass point at Yanddrawanarc.Labelthearcintersectionpointand draw an arc. Label the arc intersection pointZ.Useastraightedgetodrawray. Use a straightedge to draw ray\overrightarrow{XZ}$.

    • Result: Constructed angle \angle YXZ \cong \angle A$.\n - Constructing an Angle Bisector (Bisecting \angle Awithraywith ray\overrightarrow{AD}):\n - Definition: An angle bisector is a ray that divides an angle into two congruent adjacent angles.\n - Step 1: Place compass point at A.Drawanarcintersectingbothraysof. Draw an arc intersecting both rays of\angle A.Labelintersections. Label intersectionsBandandC$.

    • Step 2: Place compass point at BB and draw an arc in the interior of ∠A\angle A. Keeping the same compass setting (must be greater than half distance BCBC), place compass point at CC and draw an arc intersecting the arc from B$.\n - Step 3: Label the arc intersection point D.Useastraightedgetodrawray. Use a straightedge to draw ray\overrightarrow{AD}$.

    • Result: Ray AD→\overrightarrow{AD} is the angle bisector of \angle A$.\n\n- Concept Summary and Reasoning Questions\n - Constructing vs Visual Appearance: Visual assessment ("looks the same") relies on optical perception and can be inaccurate. Constructing a figure guarantees geometric equivalence (\cong) backed by mathematical postulates.\n - Practice Problem 9: Given m\angle LMN = 116^\circ,,m\angle JKM = 122^\circ,and, andm\angle JNM = 103^\circ,determine, determinem\angle NKM.\n\n# Lesson 1-3: Midpoint and Distance\n\n- Objectives and Standards\n - Learning Objective: Use the midpoint and distance formulas to solve coordinate plane problems.\n - Core Vocabulary: midpoint.\n - TEKS G.2A: Determine the coordinates of a point that is a given fractional distance less than one from one end of a line segment to the other in one- and two-dimensional coordinate systems, including finding the midpoint.\n - Mathematical Process Standards: G.1A,,G.1B,,G.1C,,G.1D,,G.1F,,G.1G$.

    • Essential Question: How are the midpoint and length of a segment on the coordinate plane determined?

  • Midpoint Formula and Derivation

    • Definition of Midpoint: The point that divides a line segment into two congruent segments.

    • Derivation:

    • The midpoint coordinates are the average of the xx-coordinates and the average of the yy-coordinates of segment endpoints P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2).

    • Midpoint Formula:       M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

    • Example 2 Calculation:

    • Find the midpoint of AB‾\overline{AB} with endpoints A(−3,2)A(-3, 2) and B(4,−2)B(4, -2):       M=(−3+42,2+(−2)2)=(12,02)=(0.5,0)M = \left( \frac{-3 + 4}{2}, \frac{2 + (-2)}{2} \right) = \left( \frac{1}{2}, \frac{0}{2} \right) = (0.5, 0)

    • Common Error Warning: Midpoint computation represents an average; add the coordinate values before dividing by 22.

    • Try It 2 Calculations:

    • Part a: Midpoint of C(−2,5)C(-2, 5) and D(8,−12)D(8, -12):       M=(−2+82,5+(−12)2)=(62,−72)=(3,−3.5)M = \left( \frac{-2 + 8}{2}, \frac{5 + (-12)}{2} \right) = \left( \frac{6}{2}, \frac{-7}{2} \right) = (3, -3.5)

    • Part b: Midpoint of E(2.5,−7)E(2.5, -7) and F(−6.2,−3.8)F(-6.2, -3.8):       M=(2.5+(−6.2)2,−7+(−3.8)2)=(−3.72,−10.82)=(−1.85,−5.4)M = \left( \frac{2.5 + (-6.2)}{2}, \frac{-7 + (-3.8)}{2} \right) = \left( \frac{-3.7}{2}, \frac{-10.8}{2} \right) = (-1.85, -5.4)

  • Partitioning a Line Segment

    • Concept: Finding coordinates for a point partitioning line segment AB‾\overline{AB} in a given ratio a:ba:b.

    • Ratio Conversion Rule: Convert part-to-part ratio a:ba:b to part-to-whole fraction aa+b\frac{a}{a+b}.

    • Example 3 Step-by-Step Procedure:

    • Task: Partition AB‾\overline{AB} in ratio 3:23:2 from A(3,−4)A(3, -4) to B(13,11)B(13, 11).

    • Step 1: Calculate fractional distance from AA to BB:       Fraction=33+2=35\text{Fraction} = \frac{3}{3 + 2} = \frac{3}{5}

    • Step 2: Compute 35\frac{3}{5} of horizontal and vertical changes:       Horizontal change (Δx)=13−3=10  ⟹  35(10)=6\text{Horizontal change } (\Delta x) = 13 - 3 = 10 \implies \frac{3}{5}(10) = 6       Vertical change (Δy)=11−(−4)=15  ⟹  35(15)=9\text{Vertical change } (\Delta y) = 11 - (-4) = 15 \implies \frac{3}{5}(15) = 9

    • Step 3: Add changes to coordinates of starting point A(3,−4)A(3, -4):       x=3+6=9x = 3 + 6 = 9       y=−4+9=5y = -4 + 9 = 5       Partitioning point coordinates=(9,5)\text{Partitioning point coordinates} = (9, 5)

    • Try It 3 Tasks:

    • Part a: Find coordinates of point partitioning AB‾\overline{AB} in ratio 7:37:3 (uses fraction 710\frac{7}{10}).

    • Part b: Find coordinates of point located 12\frac{1}{2} of the way from BB to A$.\n\n- Distance Formula and Concept Summary\n - Distance Formula:\n - Derived from Pythagorean Theorem for distance dbetweenbetweenP(x_1, y_1)andandQ(x_2, y_2):\n      d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\n - Summary Example:\n - Endpoints P(-3, 4)andandQ(1, 7).\n - Midpoint:\n      M = \left( \frac{-3 + 1}{2}, \frac{4 + 7}{2} \right) = \left( \frac{-2}{2}, \frac{11}{2} \right) = (-1, 5.5)\n - Distance:\n      d = \sqrt{(1 - (-3))^2 + (7 - 4)^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\n\n- Lesson 1-3 Practice Exercises and Solutions\n - Midpoint Formula Definition:\n    M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\n - Midpoint Computations:\n - A(-4, 6)andandB(10, -10)::M = \left( \frac{-4+10}{2}, \frac{6+(-10)}{2} \right) = (3, -2)\n - C(-3, -8)andandD(-6.5, -4.5)::M = \left( \frac{-3+(-6.5)}{2}, \frac{-8+(-4.5)}{2} \right) = (-4.75, -6.25)\n - E(3, 7)andandF(-8, -10)::M = \left( \frac{3+(-8)}{2}, \frac{7+(-10)}{2} \right) = (-2.5, -1.5)\n - G(-6, -13)andandH(-6.4, -3.8)::M = \left( \frac{-6+(-6.4)}{2}, \frac{-13+(-3.8)}{2} \right) = (-6.2, -8.4)\n - Partitioning Exercises on Segment \overline{CD}:\n - Ratio 1:2usesfractionuses fraction\frac{1}{1+2} = \frac{1}{3}oflengthfromof length fromCtotoD$.

    • Ratio 5:15:1 uses fraction 55+1=56\frac{5}{5+1} = \frac{5}{6} of length from CC to D$.\n - Point \frac{2}{3}ofthewayfromof the way fromCtotoD$.

    • Point 45\frac{4}{5} of the way from DD to C$.\n - Distance Formula Definition:\n    d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\n - Distance Computations:\n - A(6, 8)andandB(-1, 8)::d = \sqrt{(-1-6)^2 + (8-8)^2} = \sqrt{(-7)^2 + 0} = 7\n - C(5, -6)andandD(5, 6)::d = \sqrt{(5-5)^2 + (6-(-6))^2} = \sqrt{0 + 12^2} = 12\n - E(-2, 0)andandF(11, 0)::d = \sqrt{(11-(-2))^2 + (0-0)^2} = \sqrt{13^2} = 13\n - Q(1, -5)andandT(9, 1)::d = \sqrt{(9-1)^2 + (1-(-5))^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\n - Segment Midpoint Algebraic Relation:\n - If Misthemidpointofis the midpoint of\overline{ST},therelationshipbetweenfullsegment, the relationship between full segmentSTandhalfsegmentand half segmentMT is:\n      ST = 2 \cdot MT \quad \text{or} \quad MT = \frac{1}{2} ST\n - Coordinate Grid Bedroom Application:\n - Axes represent bedroom walls with one corner at origin (0,0).\n - Farthest bed corner evaluated at point (x, y) on coordinate plane.\n - Distance formula from origin: d = \sqrt{x^2 + y^2}$$.