Points, Lines, Line Segments, and Planes - Tenth Grade Geometry
Introduction to Basic Geometric Concepts
Point: A location in space with no size or dimension (represented by a dot)
Line: Extends infinitely in both directions with no endpoints
Plane: A flat surface extending infinitely in all directions
Notation: Points use capital letters (A, B, C); Lines use lowercase letters or two points
Line: Extends infinitely in both directions with no endpoints
Plane: A flat surface extending infinitely in all directions
Notation: Points use capital letters (A, B, C); Lines use lowercase letters or two points
1. Lines, Line Segments, and Rays
Line: Extends infinitely in both directions
Line Segment: Part of a line with two endpoints
Ray: Part of a line with one endpoint, extends infinitely in one direction
Collinear Points: Points that lie on the same line
Line Segment: Part of a line with two endpoints
Ray: Part of a line with one endpoint, extends infinitely in one direction
Collinear Points: Points that lie on the same line
Notation and Symbols:
Line: $\overleftrightarrow{AB}$ or line $\ell$
• Has NO endpoints
• Extends infinitely in both directions
• Arrows on both ends
Line Segment: $\overline{AB}$ or segment $AB$
• Has TWO endpoints (A and B)
• Finite length that can be measured
• No arrows
Ray: $\overrightarrow{AB}$
• Has ONE endpoint (starting point A)
• Extends infinitely in direction of B
• Arrow on one end only
• Order matters: $\overrightarrow{AB} \neq \overrightarrow{BA}$
Line: $\overleftrightarrow{AB}$ or line $\ell$
• Has NO endpoints
• Extends infinitely in both directions
• Arrows on both ends
Line Segment: $\overline{AB}$ or segment $AB$
• Has TWO endpoints (A and B)
• Finite length that can be measured
• No arrows
Ray: $\overrightarrow{AB}$
• Has ONE endpoint (starting point A)
• Extends infinitely in direction of B
• Arrow on one end only
• Order matters: $\overrightarrow{AB} \neq \overrightarrow{BA}$
Example 1: Identify and name geometric figures
Given points A, B, and C on a line:
• Line through A and B: $\overleftrightarrow{AB}$ or $\overleftrightarrow{BA}$ (same line)
• Segment from A to B: $\overline{AB}$ or $\overline{BA}$ (same segment)
• Ray starting at A through B: $\overrightarrow{AB}$ (different from $\overrightarrow{BA}$)
Given points A, B, and C on a line:
• Line through A and B: $\overleftrightarrow{AB}$ or $\overleftrightarrow{BA}$ (same line)
• Segment from A to B: $\overline{AB}$ or $\overline{BA}$ (same segment)
• Ray starting at A through B: $\overrightarrow{AB}$ (different from $\overrightarrow{BA}$)
Key Differences:
• Line: ←——————→ (infinite both ways)
• Segment: •——————• (two endpoints)
• Ray: •——————→ (one endpoint, one arrow)
• Line: ←——————→ (infinite both ways)
• Segment: •——————• (two endpoints)
• Ray: •——————→ (one endpoint, one arrow)
2. Properties of Planes, Lines, and Points
Plane: A flat surface with no thickness, extends infinitely
Coplanar: Points or lines in the same plane
Space: The set of all points in three dimensions
Coplanar: Points or lines in the same plane
Space: The set of all points in three dimensions
Fundamental Properties (Postulates):
1. Through any two points, there is exactly ONE line
• Two points determine a unique line
2. Through any three non-collinear points, there is exactly ONE plane
• Three non-collinear points determine a unique plane
3. A line contains at least two points
4. A plane contains at least three non-collinear points
5. If two points lie in a plane, then the entire line through them lies in the plane
6. If two planes intersect, their intersection is a line
1. Through any two points, there is exactly ONE line
• Two points determine a unique line
2. Through any three non-collinear points, there is exactly ONE plane
• Three non-collinear points determine a unique plane
3. A line contains at least two points
4. A plane contains at least three non-collinear points
5. If two points lie in a plane, then the entire line through them lies in the plane
6. If two planes intersect, their intersection is a line
Example 1: Apply properties
Q: How many lines can pass through points A and B?
A: Exactly ONE line (by Property 1)
Q: Do four points always determine a plane?
A: No, only if they are NOT all collinear and NOT all in the same plane already
Q: How many lines can pass through points A and B?
A: Exactly ONE line (by Property 1)
Q: Do four points always determine a plane?
A: No, only if they are NOT all collinear and NOT all in the same plane already
3. Describe Intersections in a Plane
Intersection: The set of points common to two or more geometric figures
Parallel Lines: Lines in same plane that never intersect
Intersecting Lines: Lines that meet at exactly one point
Parallel Lines: Lines in same plane that never intersect
Intersecting Lines: Lines that meet at exactly one point
Types of Intersections:
Two Lines:
• Intersect at ONE point, OR
• Are parallel (no intersection), OR
• Are the same line (infinite intersections)
Line and Plane:
• Intersect at ONE point, OR
• Line lies in the plane (infinite intersections), OR
• Are parallel (no intersection)
Two Planes:
• Intersect at a LINE, OR
• Are parallel (no intersection), OR
• Are the same plane (infinite intersections)
Two Lines:
• Intersect at ONE point, OR
• Are parallel (no intersection), OR
• Are the same line (infinite intersections)
Line and Plane:
• Intersect at ONE point, OR
• Line lies in the plane (infinite intersections), OR
• Are parallel (no intersection)
Two Planes:
• Intersect at a LINE, OR
• Are parallel (no intersection), OR
• Are the same plane (infinite intersections)
Example 1: Describe intersections
Lines $\ell$ and $m$ intersect at point P
Intersection: {P}
Planes R and S intersect along line $n$
Intersection: line $n$
Lines $\ell$ and $m$ intersect at point P
Intersection: {P}
Planes R and S intersect along line $n$
Intersection: line $n$
4. Lengths of Segments on Number Lines
Distance: The length between two points
Coordinate: The number associated with a point on a number line
Length: Always positive (non-negative)
Coordinate: The number associated with a point on a number line
Length: Always positive (non-negative)
Distance on Number Line:
For points A and B with coordinates $a$ and $b$:
$$AB = |b - a| = |a - b|$$
In words: Distance is the absolute value of the difference of coordinates
Properties:
• Distance is always ≥ 0
• $AB = BA$ (distance is same both ways)
• Use absolute value to ensure positive result
For points A and B with coordinates $a$ and $b$:
$$AB = |b - a| = |a - b|$$
In words: Distance is the absolute value of the difference of coordinates
Properties:
• Distance is always ≥ 0
• $AB = BA$ (distance is same both ways)
• Use absolute value to ensure positive result
Example 1: Find lengths on number line
Point A at -3, Point B at 5
$$AB = |5 - (-3)| = |5 + 3| = |8| = 8$$
Or: $AB = |-3 - 5| = |-8| = 8$
Answer: AB = 8 units
Point A at -3, Point B at 5
$$AB = |5 - (-3)| = |5 + 3| = |8| = 8$$
Or: $AB = |-3 - 5| = |-8| = 8$
Answer: AB = 8 units
Example 2: Find coordinate
Point C at 2, CD = 7. Find coordinate of D.
$|d - 2| = 7$
$d - 2 = 7$ OR $d - 2 = -7$
$d = 9$ OR $d = -5$
Answer: D is at 9 or -5
Point C at 2, CD = 7. Find coordinate of D.
$|d - 2| = 7$
$d - 2 = 7$ OR $d - 2 = -7$
$d = 9$ OR $d = -5$
Answer: D is at 9 or -5
5. Additive Property of Length
Segment Addition Postulate: If B is between A and C, then AB + BC = AC
Between: Point B is between A and C if A, B, and C are collinear and AB + BC = AC
Between: Point B is between A and C if A, B, and C are collinear and AB + BC = AC
Segment Addition Postulate:
If point B is between points A and C, then:
$$AB + BC = AC$$
Visual: A————B————C
Conversely: If AB + BC = AC, then B is between A and C (and A, B, C are collinear)
If point B is between points A and C, then:
$$AB + BC = AC$$
Visual: A————B————C
Conversely: If AB + BC = AC, then B is between A and C (and A, B, C are collinear)
Example 1: Use Segment Addition
Given: AB = 5, BC = 3. Find AC.
If B is between A and C:
$AC = AB + BC = 5 + 3 = 8$
Answer: AC = 8
Given: AB = 5, BC = 3. Find AC.
If B is between A and C:
$AC = AB + BC = 5 + 3 = 8$
Answer: AC = 8
Example 2: Find missing length
Given: AC = 12, AB = 7. Find BC.
$AB + BC = AC$
$7 + BC = 12$
$BC = 5$
Answer: BC = 5
Given: AC = 12, AB = 7. Find BC.
$AB + BC = AC$
$7 + BC = 12$
$BC = 5$
Answer: BC = 5
Example 3: Algebraic application
Given: AB = 2x, BC = x + 3, AC = 18. Find x.
$AB + BC = AC$
$2x + (x + 3) = 18$
$3x + 3 = 18$
$3x = 15$
$x = 5$
Answer: x = 5
Given: AB = 2x, BC = x + 3, AC = 18. Find x.
$AB + BC = AC$
$2x + (x + 3) = 18$
$3x + 3 = 18$
$3x = 15$
$x = 5$
Answer: x = 5
6. Midpoints
Midpoint: The point that divides a segment into two equal parts
Bisect: To divide into two equal parts
Segment Bisector: Any line, ray, or segment that passes through the midpoint
Bisect: To divide into two equal parts
Segment Bisector: Any line, ray, or segment that passes through the midpoint
Midpoint on Number Line:
For segment with endpoints at coordinates $a$ and $b$:
$$M = \frac{a + b}{2}$$
In words: Midpoint coordinate is the average of the endpoint coordinates
Properties of Midpoint M:
• $AM = MB$ (equal segments)
• $AM = MB = \frac{1}{2}AB$ (each half is half the whole)
• $AB = 2 \cdot AM = 2 \cdot MB$
For segment with endpoints at coordinates $a$ and $b$:
$$M = \frac{a + b}{2}$$
In words: Midpoint coordinate is the average of the endpoint coordinates
Properties of Midpoint M:
• $AM = MB$ (equal segments)
• $AM = MB = \frac{1}{2}AB$ (each half is half the whole)
• $AB = 2 \cdot AM = 2 \cdot MB$
Example 1: Find midpoint on number line
Find midpoint of segment with endpoints at -4 and 10
$$M = \frac{-4 + 10}{2} = \frac{6}{2} = 3$$
Answer: Midpoint is at 3
Find midpoint of segment with endpoints at -4 and 10
$$M = \frac{-4 + 10}{2} = \frac{6}{2} = 3$$
Answer: Midpoint is at 3
Example 2: Find endpoint
Midpoint at 5, one endpoint at 2. Find other endpoint.
$5 = \frac{2 + b}{2}$
$10 = 2 + b$
$b = 8$
Answer: Other endpoint is at 8
Midpoint at 5, one endpoint at 2. Find other endpoint.
$5 = \frac{2 + b}{2}$
$10 = 2 + b$
$b = 8$
Answer: Other endpoint is at 8
7. Weighted Averages
Weighted Average: A point that divides a segment in a specific ratio (not necessarily at midpoint)
Ratio: Comparison of two quantities
Partition Point: Point that divides segment in given ratio
Ratio: Comparison of two quantities
Partition Point: Point that divides segment in given ratio
Weighted Average Formula (1D):
To find point P that divides segment from A to B in ratio $m:n$:
$$P = \frac{n \cdot a + m \cdot b}{m + n}$$
where $a$ and $b$ are coordinates of A and B
Special Case: When $m = n$ (midpoint), $P = \frac{a+b}{2}$
To find point P that divides segment from A to B in ratio $m:n$:
$$P = \frac{n \cdot a + m \cdot b}{m + n}$$
where $a$ and $b$ are coordinates of A and B
Special Case: When $m = n$ (midpoint), $P = \frac{a+b}{2}$
Example 1: Find partition point
Find point P that divides segment from A(2) to B(12) in ratio 1:3
$$P = \frac{3(2) + 1(12)}{1 + 3} = \frac{6 + 12}{4} = \frac{18}{4} = 4.5$$
Answer: P is at 4.5
Find point P that divides segment from A(2) to B(12) in ratio 1:3
$$P = \frac{3(2) + 1(12)}{1 + 3} = \frac{6 + 12}{4} = \frac{18}{4} = 4.5$$
Answer: P is at 4.5
8. Congruent Line Segments
Congruent Segments: Segments with equal lengths
Symbol: $\cong$ (read as "is congruent to")
Notation: $\overline{AB} \cong \overline{CD}$ means AB = CD
Symbol: $\cong$ (read as "is congruent to")
Notation: $\overline{AB} \cong \overline{CD}$ means AB = CD
Congruence Definition:
$$\overline{AB} \cong \overline{CD} \iff AB = CD$$
In words: Segments are congruent if and only if they have equal lengths
Properties of Congruence:
• Reflexive: $\overline{AB} \cong \overline{AB}$
• Symmetric: If $\overline{AB} \cong \overline{CD}$, then $\overline{CD} \cong \overline{AB}$
• Transitive: If $\overline{AB} \cong \overline{CD}$ and $\overline{CD} \cong \overline{EF}$, then $\overline{AB} \cong \overline{EF}$
$$\overline{AB} \cong \overline{CD} \iff AB = CD$$
In words: Segments are congruent if and only if they have equal lengths
Properties of Congruence:
• Reflexive: $\overline{AB} \cong \overline{AB}$
• Symmetric: If $\overline{AB} \cong \overline{CD}$, then $\overline{CD} \cong \overline{AB}$
• Transitive: If $\overline{AB} \cong \overline{CD}$ and $\overline{CD} \cong \overline{EF}$, then $\overline{AB} \cong \overline{EF}$
Example 1: Identify congruent segments
Given: AB = 5 cm, CD = 5 cm, EF = 7 cm
$\overline{AB} \cong \overline{CD}$ (both have length 5)
$\overline{AB} \not\cong \overline{EF}$ (different lengths)
$\overline{CD} \not\cong \overline{EF}$ (different lengths)
Given: AB = 5 cm, CD = 5 cm, EF = 7 cm
$\overline{AB} \cong \overline{CD}$ (both have length 5)
$\overline{AB} \not\cong \overline{EF}$ (different lengths)
$\overline{CD} \not\cong \overline{EF}$ (different lengths)
9. Construct a Congruent Segment
Steps to Construct Congruent Segment (with compass and straightedge):
Given: Segment $\overline{AB}$
Goal: Construct $\overline{CD} \cong \overline{AB}$
Step 1: Draw a ray with endpoint C
Step 2: Set compass width to length AB
Step 3: Place compass point on C
Step 4: Draw an arc intersecting the ray at point D
Step 5: $\overline{CD} \cong \overline{AB}$
Given: Segment $\overline{AB}$
Goal: Construct $\overline{CD} \cong \overline{AB}$
Step 1: Draw a ray with endpoint C
Step 2: Set compass width to length AB
Step 3: Place compass point on C
Step 4: Draw an arc intersecting the ray at point D
Step 5: $\overline{CD} \cong \overline{AB}$
10. Perpendicular Bisector Theorem
Perpendicular Bisector: A line that passes through the midpoint of a segment and is perpendicular to it
Perpendicular: Forms 90° angles
Equidistant: Same distance
Perpendicular: Forms 90° angles
Equidistant: Same distance
Perpendicular Bisector Theorem:
Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If point P is on perpendicular bisector of $\overline{AB}$, then:
$$PA = PB$$
Converse: If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment.
If $PA = PB$, then point P is on perpendicular bisector of $\overline{AB}$
Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If point P is on perpendicular bisector of $\overline{AB}$, then:
$$PA = PB$$
Converse: If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment.
If $PA = PB$, then point P is on perpendicular bisector of $\overline{AB}$
Example 1: Apply theorem
Line $\ell$ is perpendicular bisector of $\overline{AB}$. Point P is on line $\ell$. If PA = 8, find PB.
By Perpendicular Bisector Theorem:
$PA = PB$
Therefore, $PB = 8$
Answer: PB = 8
Line $\ell$ is perpendicular bisector of $\overline{AB}$. Point P is on line $\ell$. If PA = 8, find PB.
By Perpendicular Bisector Theorem:
$PA = PB$
Therefore, $PB = 8$
Answer: PB = 8
11. Midpoint Formula: Find the Midpoint
Coordinate Plane: Two-dimensional plane with x and y axes
Ordered Pair: (x, y) represents location of point
Midpoint in 2D: Point halfway between two points
Ordered Pair: (x, y) represents location of point
Midpoint in 2D: Point halfway between two points
Midpoint Formula (2D):
For points $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
In words: Average the x-coordinates, average the y-coordinates
Components:
• x-coordinate of M: $\frac{x_1 + x_2}{2}$
• y-coordinate of M: $\frac{y_1 + y_2}{2}$
For points $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
In words: Average the x-coordinates, average the y-coordinates
Components:
• x-coordinate of M: $\frac{x_1 + x_2}{2}$
• y-coordinate of M: $\frac{y_1 + y_2}{2}$
Example 1: Find midpoint
Find midpoint of segment with endpoints A(2, 5) and B(8, 11)
$$M = \left(\frac{2 + 8}{2}, \frac{5 + 11}{2}\right)$$
$$M = \left(\frac{10}{2}, \frac{16}{2}\right) = (5, 8)$$
Answer: M(5, 8)
Find midpoint of segment with endpoints A(2, 5) and B(8, 11)
$$M = \left(\frac{2 + 8}{2}, \frac{5 + 11}{2}\right)$$
$$M = \left(\frac{10}{2}, \frac{16}{2}\right) = (5, 8)$$
Answer: M(5, 8)
Example 2: Midpoint with negative coordinates
Find midpoint of C(-3, 4) and D(5, -2)
$$M = \left(\frac{-3 + 5}{2}, \frac{4 + (-2)}{2}\right)$$
$$M = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)$$
Answer: M(1, 1)
Find midpoint of C(-3, 4) and D(5, -2)
$$M = \left(\frac{-3 + 5}{2}, \frac{4 + (-2)}{2}\right)$$
$$M = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)$$
Answer: M(1, 1)
12. Partition a Line Segment in a Given Ratio
Partition Point: Point that divides segment in specific ratio
Directed Segment: Segment with specified direction from start to end
Ratio m:n: Point is $\frac{m}{m+n}$ of way from start to end
Directed Segment: Segment with specified direction from start to end
Ratio m:n: Point is $\frac{m}{m+n}$ of way from start to end
Section Formula (Partition in Ratio m:n):
For points $A(x_1, y_1)$ and $B(x_2, y_2)$, find point P that divides $\overline{AB}$ in ratio $m:n$:
$$P = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right)$$
Alternative (using fraction c = m/(m+n)):
$$P = (x_1 + c(x_2 - x_1), y_1 + c(y_2 - y_1))$$
where $c = \frac{m}{m + n}$ (fraction of distance from A to B)
For points $A(x_1, y_1)$ and $B(x_2, y_2)$, find point P that divides $\overline{AB}$ in ratio $m:n$:
$$P = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right)$$
Alternative (using fraction c = m/(m+n)):
$$P = (x_1 + c(x_2 - x_1), y_1 + c(y_2 - y_1))$$
where $c = \frac{m}{m + n}$ (fraction of distance from A to B)
Example 1: Partition in ratio 2:3
Find point P that divides segment from A(1, 2) to B(6, 12) in ratio 2:3
$m = 2$, $n = 3$
$$P_x = \frac{2(6) + 3(1)}{2 + 3} = \frac{12 + 3}{5} = \frac{15}{5} = 3$$
$$P_y = \frac{2(12) + 3(2)}{2 + 3} = \frac{24 + 6}{5} = \frac{30}{5} = 6$$
Answer: P(3, 6)
Find point P that divides segment from A(1, 2) to B(6, 12) in ratio 2:3
$m = 2$, $n = 3$
$$P_x = \frac{2(6) + 3(1)}{2 + 3} = \frac{12 + 3}{5} = \frac{15}{5} = 3$$
$$P_y = \frac{2(12) + 3(2)}{2 + 3} = \frac{24 + 6}{5} = \frac{30}{5} = 6$$
Answer: P(3, 6)
Example 2: Using fraction method
Find point 3/4 of way from A(0, 0) to B(8, 12)
$c = \frac{3}{4}$
$$P_x = 0 + \frac{3}{4}(8 - 0) = \frac{3}{4}(8) = 6$$
$$P_y = 0 + \frac{3}{4}(12 - 0) = \frac{3}{4}(12) = 9$$
Answer: P(6, 9)
Find point 3/4 of way from A(0, 0) to B(8, 12)
$c = \frac{3}{4}$
$$P_x = 0 + \frac{3}{4}(8 - 0) = \frac{3}{4}(8) = 6$$
$$P_y = 0 + \frac{3}{4}(12 - 0) = \frac{3}{4}(12) = 9$$
Answer: P(6, 9)
13. Midpoint Formula: Find the Endpoint
Finding Missing Endpoint:
Given: Midpoint $M(x_m, y_m)$ and one endpoint $A(x_1, y_1)$
Find: Other endpoint $B(x_2, y_2)$
From midpoint formula:
$$x_m = \frac{x_1 + x_2}{2} \implies x_2 = 2x_m - x_1$$
$$y_m = \frac{y_1 + y_2}{2} \implies y_2 = 2y_m - y_1$$
Formula: $B(2x_m - x_1, 2y_m - y_1)$
Given: Midpoint $M(x_m, y_m)$ and one endpoint $A(x_1, y_1)$
Find: Other endpoint $B(x_2, y_2)$
From midpoint formula:
$$x_m = \frac{x_1 + x_2}{2} \implies x_2 = 2x_m - x_1$$
$$y_m = \frac{y_1 + y_2}{2} \implies y_2 = 2y_m - y_1$$
Formula: $B(2x_m - x_1, 2y_m - y_1)$
Example 1: Find missing endpoint
Midpoint M(3, 5), one endpoint A(1, 2). Find endpoint B.
$$x_2 = 2(3) - 1 = 6 - 1 = 5$$
$$y_2 = 2(5) - 2 = 10 - 2 = 8$$
Answer: B(5, 8)
Midpoint M(3, 5), one endpoint A(1, 2). Find endpoint B.
$$x_2 = 2(3) - 1 = 6 - 1 = 5$$
$$y_2 = 2(5) - 2 = 10 - 2 = 8$$
Answer: B(5, 8)
Example 2: Verify answer
From Example 1, check: Is M midpoint of A(1,2) and B(5,8)?
$$M = \left(\frac{1+5}{2}, \frac{2+8}{2}\right) = \left(\frac{6}{2}, \frac{10}{2}\right) = (3, 5)$$ ✓
From Example 1, check: Is M midpoint of A(1,2) and B(5,8)?
$$M = \left(\frac{1+5}{2}, \frac{2+8}{2}\right) = \left(\frac{6}{2}, \frac{10}{2}\right) = (3, 5)$$ ✓
14. Distance Formula
Distance: Length between two points in coordinate plane
Derived from: Pythagorean Theorem
Always positive: Distance ≥ 0
Derived from: Pythagorean Theorem
Always positive: Distance ≥ 0
Distance Formula (2D):
For points $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Derivation from Pythagorean Theorem:
• Horizontal distance: $|x_2 - x_1|$
• Vertical distance: $|y_2 - y_1|$
• Hypotenuse (distance): $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$
Note: Order doesn't matter: $d(A,B) = d(B,A)$
For points $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Derivation from Pythagorean Theorem:
• Horizontal distance: $|x_2 - x_1|$
• Vertical distance: $|y_2 - y_1|$
• Hypotenuse (distance): $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$
Note: Order doesn't matter: $d(A,B) = d(B,A)$
Example 1: Find distance
Find distance between A(1, 2) and B(4, 6)
$$d = \sqrt{(4-1)^2 + (6-2)^2}$$
$$d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: d = 5 units
Find distance between A(1, 2) and B(4, 6)
$$d = \sqrt{(4-1)^2 + (6-2)^2}$$
$$d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: d = 5 units
Example 2: Distance with negative coordinates
Find distance between C(-2, 3) and D(1, -1)
$$d = \sqrt{(1-(-2))^2 + (-1-3)^2}$$
$$d = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: d = 5 units
Find distance between C(-2, 3) and D(1, -1)
$$d = \sqrt{(1-(-2))^2 + (-1-3)^2}$$
$$d = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: d = 5 units
Example 3: Irrational distance
Find distance between E(0, 0) and F(3, 4)
$$d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
But for E(0,0) and G(2, 3):
$$d = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$$
Find distance between E(0, 0) and F(3, 4)
$$d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
But for E(0,0) and G(2, 3):
$$d = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.61$$
15. Distance to the Origin in Three Dimensions
3D Coordinate System: Uses three axes (x, y, z)
Origin: Point (0, 0, 0)
Three-dimensional point: (x, y, z)
Origin: Point (0, 0, 0)
Three-dimensional point: (x, y, z)
Distance from Origin in 3D:
For point $P(x, y, z)$:
$$d = \sqrt{x^2 + y^2 + z^2}$$
General 3D Distance Formula:
For points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$:
$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
For point $P(x, y, z)$:
$$d = \sqrt{x^2 + y^2 + z^2}$$
General 3D Distance Formula:
For points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$:
$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
Example 1: Distance from origin
Find distance from origin to P(3, 4, 5)
$$d = \sqrt{3^2 + 4^2 + 5^2}$$
$$d = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \approx 7.07$$
Answer: $d = 5\sqrt{2}$ units
Find distance from origin to P(3, 4, 5)
$$d = \sqrt{3^2 + 4^2 + 5^2}$$
$$d = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \approx 7.07$$
Answer: $d = 5\sqrt{2}$ units
Example 2: 3D distance between points
Find distance between A(1, 2, 3) and B(4, 6, 8)
$$d = \sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2}$$
$$d = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$$
Answer: $d = 5\sqrt{2}$ units
Find distance between A(1, 2, 3) and B(4, 6, 8)
$$d = \sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2}$$
$$d = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$$
Answer: $d = 5\sqrt{2}$ units
16. Construct the Midpoint or Perpendicular Bisector
Construction: Using only compass and straightedge (no measurements)
Perpendicular Bisector: Line perpendicular to segment at its midpoint
Perpendicular Bisector: Line perpendicular to segment at its midpoint
Construction: Perpendicular Bisector of Segment AB
Step 1: Place compass point on A
Step 2: Set compass width to more than half of AB
Step 3: Draw an arc above and below AB
Step 4: Keep same compass width, place compass on B
Step 5: Draw arcs intersecting the first arcs (create two points P and Q)
Step 6: Draw line through P and Q
Result: Line PQ is perpendicular bisector of AB
Bonus: Intersection of PQ and AB is the midpoint M
Step 1: Place compass point on A
Step 2: Set compass width to more than half of AB
Step 3: Draw an arc above and below AB
Step 4: Keep same compass width, place compass on B
Step 5: Draw arcs intersecting the first arcs (create two points P and Q)
Step 6: Draw line through P and Q
Result: Line PQ is perpendicular bisector of AB
Bonus: Intersection of PQ and AB is the midpoint M
Why This Works:
• Points P and Q are equidistant from A and B
• By Perpendicular Bisector Theorem (converse), P and Q lie on perpendicular bisector
• Two points determine a unique line (the perpendicular bisector)
• Points P and Q are equidistant from A and B
• By Perpendicular Bisector Theorem (converse), P and Q lie on perpendicular bisector
• Two points determine a unique line (the perpendicular bisector)
Notation Summary
| Figure | Notation | Read As | Endpoints | Arrows |
|---|---|---|---|---|
| Point | $A$ | Point A | None (just location) | None |
| Line | $\overleftrightarrow{AB}$ or $\ell$ | Line AB or line ell | None (infinite) | Both ends |
| Segment | $\overline{AB}$ or $AB$ | Segment AB | Two (A and B) | None |
| Ray | $\overrightarrow{AB}$ | Ray AB | One (A is endpoint) | One end (toward B) |
| Plane | Plane $P$ or plane $ABC$ | Plane P | None (infinite) | N/A |
Formula Quick Reference
| Formula Name | Formula | Use |
|---|---|---|
| Distance (1D) | $d = |b - a|$ | Distance between points on number line |
| Distance (2D) | $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ | Distance between points in plane |
| Distance (3D) | $d = \sqrt{x^2 + y^2 + z^2}$ | Distance from origin in space |
| Midpoint (1D) | $M = \frac{a+b}{2}$ | Midpoint on number line |
| Midpoint (2D) | $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ | Midpoint in coordinate plane |
| Segment Addition | $AB + BC = AC$ | When B is between A and C |
| Partition (ratio m:n) | $P = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)$ | Divide segment in given ratio |
| Find Endpoint | $B(2x_m - x_1, 2y_m - y_1)$ | Given midpoint and one endpoint |
Key Properties and Postulates
| Property/Postulate | Statement |
|---|---|
| Two Point Postulate | Through any two points, there is exactly one line |
| Three Point Postulate | Through any three non-collinear points, there is exactly one plane |
| Segment Addition | If B is between A and C, then $AB + BC = AC$ |
| Perpendicular Bisector Theorem | Point on perp. bisector is equidistant from endpoints |
| Converse of Perp. Bis. Theorem | Point equidistant from endpoints lies on perp. bisector |
| Midpoint Definition | M is midpoint of AB if $AM = MB = \frac{1}{2}AB$ |
Success Tips for Points, Lines, and Planes:
✓ Line has arrows on BOTH ends; Ray has arrow on ONE end; Segment has NO arrows
✓ For distance, always use absolute value or square root (always positive)
✓ Midpoint formula: AVERAGE the coordinates
✓ Distance formula comes from Pythagorean Theorem
✓ Segment Addition: If B between A and C, then AB + BC = AC
✓ To find missing endpoint: Double midpoint coordinates, subtract known endpoint
✓ Partition in ratio m:n: Point is $\frac{m}{m+n}$ of way from start
✓ Perpendicular bisector creates two equal segments at 90°
✓ Congruent segments: $\overline{AB} \cong \overline{CD}$ means AB = CD
✓ Practice with both positive and negative coordinates!
✓ Line has arrows on BOTH ends; Ray has arrow on ONE end; Segment has NO arrows
✓ For distance, always use absolute value or square root (always positive)
✓ Midpoint formula: AVERAGE the coordinates
✓ Distance formula comes from Pythagorean Theorem
✓ Segment Addition: If B between A and C, then AB + BC = AC
✓ To find missing endpoint: Double midpoint coordinates, subtract known endpoint
✓ Partition in ratio m:n: Point is $\frac{m}{m+n}$ of way from start
✓ Perpendicular bisector creates two equal segments at 90°
✓ Congruent segments: $\overline{AB} \cong \overline{CD}$ means AB = CD
✓ Practice with both positive and negative coordinates!