Study guide
7.3 Coordinate Geometry: Slope, Distance, Midpoint & Linear Graphs
Armed Forces Qualification Test AFQT
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1 Chapter 1: AFQT Structure, Scoring & Enlistment Requirements
1.1 AFQT Overview, Subtest Breakdown & CAT-ASVAB vs. P&P Format 1.2 The AFQT Scoring Formula (2×VE + AR + MK) & Percentile Norming 1.3 Branch Minimum Scores, AFQT Categories & Retake Policies
2 Chapter 2: Arithmetic Reasoning (AR) — Ratios, Rates & Percentages
2.1 Ratios, Proportions & Scale Conversions 2.2 Percentages, Discounts, Interest & Tax Calculations 2.3 Rate, Distance, Time & Speed Word Problems 2.4 Work Rates & Combined Labor Problems
3 Chapter 3: Arithmetic Reasoning (AR) — Mixtures, Sequences & Applied Math
3.1 Mixture, Concentration & Price-Blending Problems 3.2 Sequences, Patterns & Series Calculations 3.3 Averages, Weighted Averages & Statistical Word Problems
4 Chapter 4: Word Knowledge (WK) — Vocabulary & Word Structure
4.1 High-Frequency AFQT Vocabulary & Synonym Recognition 4.2 Latin & Greek Roots, Prefixes & Suffixes for Word Decoding 4.3 Context Clues & Words with Multiple Meanings
5 Chapter 5: Paragraph Comprehension (PC) — Reading Analysis & Inference
5.1 Identifying Main Ideas & Primary Purpose 5.2 Locating Supporting Details & Direct Facts 5.3 Drawing Logical Inferences & Implied Conclusions 5.4 Analyzing Tone, Author Perspective & Contextual Vocabulary
6 Chapter 6: Mathematics Knowledge (MK) — Fundamentals & Linear Algebra
6.1 Arithmetic Operations, Order of Operations & Number Properties 6.2 Exponents, Scientific Notation & Square Roots / Radicals 6.3 Linear Equations, Inequalities & Absolute Value 6.4 Systems of Linear Equations & Word-to-Algebra Translation
7 Chapter 7: Mathematics Knowledge (MK) — Advanced Algebra & Functions
7.1 Polynomial Operations & Factoring Techniques 7.2 Quadratic Equations, Factoring & Quadratic Formula 7.3 Coordinate Geometry: Slope, Distance, Midpoint & Linear Graphs
8 Chapter 8: Mathematics Knowledge (MK) — Geometry & Spatial Math
8.1 Angles, Triangles & Pythagorean Theorem 8.2 Quadrilaterals, Polygons & Circle Geometry 8.3 3D Geometry: Surface Area & Volume of Solids 8.4 Basic Geometric Transformations & Trigonometric Ratios
7.3 Coordinate Geometry: Slope, Distance, Midpoint & Linear Graphs
Key Takeaways Explain This Topic AI
The slope formula m = (y2 - y1) / (x2 - x1) measures the steepness and direction of a line through two points (x1, y1) and (x2, y2).
Parallel lines have equal slopes (m1 = m2), whereas perpendicular lines have negative reciprocal slopes (m1 * m2 = -1).
The slope-intercept form y = mx + b identifies the slope m and the y-intercept (0, b) directly from the linear equation.
The distance formula d = √((x2 - x1)^2 + (y2 - y1)^2) calculates the straight-line distance between two points on the Cartesian plane.
The midpoint formula M = ((x1 + x2)/2, (y1 + y2)/2) calculates the exact coordinates of the middle point connecting two coordinate endpoints.
Last updated: July 2026
7.3 Coordinate Geometry: Slope, Distance, Midpoint & Linear Graphs
Coordinate geometry bridges algebra and geometry by representing geometric shapes and linear relationships on a two-dimensional grid. On the AFQT Mathematics Knowledge subtest, coordinate geometry questions test your ability to work with linear equations, calculate slope, determine line intersections, and compute distance and midpoint values between coordinate pairs.
1. The Cartesian Coordinate Plane
The Cartesian coordinate plane is formed by two perpendicular number lines that intersect at the origin ( 0 , 0 ) (0,0) ( 0 , 0 ) :
Horizontal axis : x x x -axis (positive to the right, negative to the left).
Vertical axis : y y y -axis (positive upward, negative downward).
The plane is divided into four quadrants , numbered counterclockwise:
- Quadrant I : ( + x , + y ) (+x, +y) ( + x , + y ) — Top-right
- Quadrant II : ( − x , + y ) (-x, +y) ( − x , + y ) — Top-left
- Quadrant III : ( − x , − y ) (-x, -y) ( − x , − y ) — Bottom-left
- Quadrant IV : ( + x , − y ) (+x, -y) ( + x , − y ) — Bottom-right
Points are written as ordered pairs ( x , y ) (x, y) ( x , y ) , where x x x represents the horizontal displacement from the origin and y y y represents the vertical displacement.
2. Slope of a Line ( m m m )
The slope of a line measures its steepness and direction. It is defined as the ratio of vertical change ( rise ) to horizontal change ( run ) between any two distinct points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) on the line: m = Rise Run = y 2 − y 1 x 2 − x 1 m = \frac{\text{Rise}}{\text{Run}} = \frac{y_2 - y_1}{x_2 - x_1} m = Run Rise = x 2 − x 1 y 2 − y 1
The Four Types of Slope
- Slope Value Visual Orientation Behavior
- Positive ( m > 0 m > 0 m > 0 ) Slanted UP from left to right As x x x increases, y y y increases
- Negative ( m < 0 m < 0 m < 0 ) Slanted DOWN from left to right As x x x increases, y y y decreases
- Zero ( m = 0 m = 0 m = 0 ) Horizontal line y y y is constant ( y = c y = c y = c ); no vertical change
Undefined ( m = undefined m = \text{undefined} m = undefined ) Vertical line x x x is constant ( x = c x = c x = c ); division by zero ( x 2 − x 1 = 0 x_2 - x_1 = 0 x 2 − x 1 = 0 )
Worked Example 1: Calculating Slope
Find the slope of the line passing through points A ( − 3 , 4 ) A(-3, 4) A ( − 3 , 4 ) and B ( 5 , − 2 ) B(5, -2) B ( 5 , − 2 ) .
Label coordinates: ( x 1 , y 1 ) = ( − 3 , 4 ) (x_1, y_1) = (-3, 4) ( x 1 , y 1 ) = ( − 3 , 4 ) and ( x 2 , y 2 ) = ( 5 , − 2 ) (x_2, y_2) = (5, -2) ( x 2 , y 2 ) = ( 5 , − 2 ) .
Apply the slope formula: m = − 2 − 4 5 − ( − 3 ) = − 6 5 + 3 = − 6 8 = − 3 4 m = \frac{-2 - 4}{5 - (-3)} = \frac{-6}{5 + 3} = \frac{-6}{8} = -\frac{3}{4} m = 5 − ( − 3 ) − 2 − 4 = 5 + 3 − 6 = 8 − 6 = − 4 3
The slope is − 3 4 -\frac{3}{4} − 4 3 (a negative slope slanting downward).
3. Parallel and Perpendicular Lines
Analyzing relationships between two lines is a frequent AFQT question type.
Parallel Lines
Two non-vertical lines are parallel if and only if they have the exact same slope : m 1 = m 2 m_1 = m_2 m 1 = m 2
Parallel lines never intersect.
Perpendicular Lines
Two non-vertical lines are perpendicular if and only if their slopes are opposite reciprocals (their product equals − 1 -1 − 1 ): m 1 ⋅ m 2 = − 1 ⟹ m 2 = − 1 m 1 m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1} m 1 ⋅ m 2 = − 1 ⟹ m 2 = − m 1 1
Perpendicular lines intersect at a right angle ( 90 ∘ 90^\circ 9 0 ∘ ).
Worked Example 2: Parallel and Perpendicular Slopes
Given Line 1 with equation y = 2 5 x + 7 y = \frac{2}{5}x + 7 y = 5 2 x + 7 :
Any line parallel to Line 1 has slope m = 2 5 m = \mathbf{\frac{2}{5}} m = 5 2 .
Any line perpendicular to Line 1 has slope m = − 5 2 m = \mathbf{-\frac{5}{2}} m = − 2 5 .
4. Forms of Linear Equations
Linear equations can be expressed in three primary standard formats on the AFQT.
- 1. Slope-Intercept Form ( y = m x + b y = mx + b y = m x + b )
- This is the most useful form for graphing and identifying line properties: y = m x + b y = mx + b y = m x + b
m m m : Slope of the line.
b b b : y y y -intercept (the point where the line crosses the y y y -axis, ( 0 , b ) (0, b) ( 0 , b ) ).
Finding Intercepts:
y y y -intercept : Set x = 0 x = 0 x = 0 and solve for y y y (equals b b b ).
x x x -intercept : Set y = 0 y = 0 y = 0 and solve for x x x (equals − b m -\frac{b}{m} − m b ).
2. Point-Slope Form ( y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) )
Used to construct the equation of a line when given a point ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and the slope m m m : y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 )
Worked Example 3: Writing a Line Equation
Find the equation of the line passing through point ( 4 , − 1 ) (4, -1) ( 4 , − 1 ) with slope m = 3 m = 3 m = 3 , in slope-intercept form.
Substitute into point-slope form: y − ( − 1 ) = 3 ( x − 4 ) ⟹ y + 1 = 3 x − 12 y - (-1) = 3(x - 4) \implies y + 1 = 3x - 12 y − ( − 1 ) = 3 ( x − 4 ) ⟹ y + 1 = 3 x − 12
Isolate y y y : y = 3 x − 13 y = 3x - 13 y = 3 x − 13
The slope-intercept form is y = 3 x − 13 y = 3x - 13 y = 3 x − 13 .
3. Standard Form ( A x + B y = C Ax + By = C A x + B y = C )
Where A , B , C A, B, C A , B , C are integers (preferably A ≥ 0 A \geq 0 A ≥ 0 ).
Slope m = − A B m = -\frac{A}{B} m = − B A .
y y y -intercept = C B = \frac{C}{B} = B C .
5. The Distance Formula
The distance formula calculates the straight-line distance d d d between two points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) on the coordinate plane. It is derived directly from the Pythagorean Theorem ( a 2 + b 2 = c 2 a^2 + b^2 = c^2 a 2 + b 2 = c 2 ): d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
Worked Example 4: Calculating Distance
Find the distance between points P ( 1 , 2 ) P(1, 2) P ( 1 , 2 ) and Q ( 7 , 10 ) Q(7, 10) Q ( 7 , 10 ) .
Assign coordinates: ( x 1 , y 1 ) = ( 1 , 2 ) (x_1, y_1) = (1, 2) ( x 1 , y 1 ) = ( 1 , 2 ) and ( x 2 , y 2 ) = ( 7 , 10 ) (x_2, y_2) = (7, 10) ( x 2 , y 2 ) = ( 7 , 10 ) .
Calculate horizontal change ( Δ x \Delta x Δ x ) and vertical change ( Δ y \Delta y Δ y ): x 2 − x 1 = 7 − 1 = 6 x_2 - x_1 = 7 - 1 = 6 x 2 − x 1 = 7 − 1 = 6 y 2 − y 1 = 10 − 2 = 8 y_2 - y_1 = 10 - 2 = 8 y 2 − y 1 = 10 − 2 = 8
Substitute into distance formula: d = 6 2 + 8 2 = 36 + 64 = 100 = 10 d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 d = 6 2 + 8 2 = 36 + 64 = 100 = 10
The distance is 10 10 10 units .
6. The Midpoint Formula
The midpoint formula finds the point M ( x m , y m ) M(x_m, y_m) M ( x m , y m ) located exactly halfway between two endpoints ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) . It is simply the average of the x x x -coordinates and the average of the y y y -coordinates: M = ( x 1 + x 2 2 , y 1 + y 2 2 ) M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) M = ( 2 x 1 + x 2 , 2 y 1 + y 2 )
Worked Example 5: Finding the Midpoint
Find the midpoint M M M of the segment connecting A ( − 6 , 8 ) A(-6, 8) A ( − 6 , 8 ) and B ( 4 , 2 ) B(4, 2) B ( 4 , 2 ) .
Average the x x x -coordinates: x m = − 6 + 4 2 = − 2 2 = − 1 x_m = \frac{-6 + 4}{2} = \frac{-2}{2} = -1 x m = 2 − 6 + 4 = 2 − 2 = − 1
Average the y y y -coordinates: y m = 8 + 2 2 = 10 2 = 5 y_m = \frac{8 + 2}{2} = \frac{10}{2} = 5 y m = 2 8 + 2 = 2 10 = 5
The midpoint coordinates are M ( − 1 , 5 ) M(-1, 5) M ( − 1 , 5 ) .
Worked Example 6: Finding an Endpoint Given Midpoint
The midpoint of segment C D CD C D is M ( 3 , 4 ) M(3, 4) M ( 3 , 4 ) . If endpoint C C C is at ( 1 , 1 ) (1, 1) ( 1 , 1 ) , find the coordinates of endpoint D ( x 2 , y 2 ) D(x_2, y_2) D ( x 2 , y 2 ) .
Use midpoint x x x -equation: 1 + x 2 2 = 3 ⟹ 1 + x 2 = 6 ⟹ x 2 = 5 \frac{1 + x_2}{2} = 3 \implies 1 + x_2 = 6 \implies x_2 = 5 2 1 + x 2 = 3 ⟹ 1 + x 2 = 6 ⟹ x 2 = 5
Use midpoint y y y -equation: 1 + y 2 2 = 4 ⟹ 1 + y 2 = 8 ⟹ y 2 = 7 \frac{1 + y_2}{2} = 4 \implies 1 + y_2 = 8 \implies y_2 = 7 2 1 + y 2 = 4 ⟹ 1 + y 2 = 8 ⟹ y 2 = 7
Endpoint D D D is at ( 5 , 7 ) (5, 7) ( 5 , 7 ) .
7. Summary of Key Formulas for AFQT Coordinate Geometry
Concept Formula Purpose
Slope m = y 2 − y 1 x 2 − x 1 m = \frac{y_2 - y_1}{x_2 - x_1} m = x 2 − x 1 y 2 − y 1 Measures line steepness and direction
Slope-Intercept Form y = m x + b y = mx + b y = m x + b Line equation with slope m m m and y y y -intercept b b b
Distance Formula d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 Length of line segment between two points
Midpoint Formula M = ( x 1 + x 2 2 , y 1 + y 2 2 ) M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) M = ( 2 x 1 + x 2 , 2 y 1 + y 2 ) Center point between two coordinates
Perpendicular Slopes m 1 ⋅ m 2 = − 1 m_1 \cdot m_2 = -1 m 1 ⋅ m 2 = − 1 Negative reciprocal slopes intersect at 90 ∘ 90^\circ 9 0 ∘
8. AFQT Tips & Avoidable Mistakes
Coordinate Order Consistency: When calculating slope m = y 2 − y 1 x 2 − x 1 m = \frac{y_2 - y_1}{x_2 - x_1} m = x 2 − x 1 y 2 − y 1 , make sure the same point's coordinates come first in both the numerator and denominator.
Minus Sign Traps in Distance: Subtracting negative coordinates produces addition: ( x 2 − ( − 3 ) ) 2 = ( x 2 + 3 ) 2 (x_2 - (-3))^2 = (x_2 + 3)^2 ( x 2 − ( − 3 ) ) 2 = ( x 2 + 3 ) 2 .
Pythagorean Triples Shortcut: Notice that distance calculation 6 2 + 8 2 = 10 \sqrt{6^2 + 8^2} = 10 6 2 + 8 2 = 10 is based on the 3-4-5 right triangle ratio (scaled by 2). Recognizing standard triples ( 3 - 4 - 5 3\text{-}4\text{-}5 3 - 4 - 5 , 5 - 12 - 13 5\text{-}12\text{-}13 5 - 12 - 13 , 8 - 15 - 17 8\text{-}15\text{-}17 8 - 15 - 17 ) saves precious calculation time on the test!