HSG.GPE.B.5: Proving and Using the Slope Criteria for Parallel and Perpendicular Lines
In plain English: HSG.GPE.B.5 is the Common Core geometry standard that asks students to prove the slope criteria for parallel and perpendicular lines and use them to solve problems. Students show why parallel lines have equal slopes and perpendicular lines have slopes with a product of -1, then write equations of parallel and perpendicular lines through given points. It is usually taught in Geometry.
Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).
Common Core State Standards for Mathematics · Domain: Expressing Geometric Properties with Equations (GPE) · Cluster: Use coordinates to prove simple geometric theorems algebraically Also written as HSG-GPE.B.5 or G-GPE.5 · Official standard
Students prove the two slope criteria and then use them. For parallel lines, an algebraic argument shows that two distinct non-vertical lines have no common point exactly when their slopes are equal. For perpendicular lines, the Pythagorean Theorem and its converse, applied to the points (1, m₁) and (1, m₂) on two lines through the origin, show that the lines meet at a right angle exactly when m₁m₂ = -1. A rotation of a slope triangle gives a second, visual argument.
With the criteria proved, students use them to solve geometric problems: they write the equation of a line parallel or perpendicular to a given line through a given point, find altitudes and perpendicular bisectors, find a missing vertex of a parallelogram, and decide whether a figure has parallel or perpendicular sides. The lesson also treats the special case of horizontal and vertical lines.
Learning Objectives
By the end of this lesson, students will be able to:
Prove that two distinct non-vertical lines are parallel if and only if their slopes are equal
Prove that two non-vertical lines are perpendicular if and only if the product of their slopes is -1
Write the equation of a line parallel or perpendicular to a given line through a given point, including horizontal and vertical cases
Use the slope criteria to solve geometric problems such as altitudes, perpendicular bisectors and missing vertices
Prior Knowledge Required
Students should already be comfortable with:
Slope as rise over run and the equation y = mx + b 8.EE.B.6
The Pythagorean Theorem and its converse 8.G.B.6
Solving linear equations and rewriting equations such as 2x + 5y = 10 in slope-intercept form HSA.REI.B.3
Systems of two linear equations and what "no solution" means HSA.REI.C.6
Rotations and translations on the coordinate plane HSG.CO.A.5
Ask students to graph three lines on one grid: line p through (0, 1) and (3, 3), line q through (0, -2) and (3, 0), and line r through (0, 4) and (2, 1).
Warm-Up Prompt
"Find the slope of each line. Which lines look parallel? Which look perpendicular? What do you notice about their slopes, and do you think the pattern always holds?"
Record the slopes: p and q both have slope 2/3, and r has slope -3/2. Students usually notice "same slope, parallel" quickly. For r, draw out the pattern "flip and change the sign." Then ask the key question of the lesson: why should this be true for every pair of lines, and not only for the ones we drew? That is what the class will prove.
Direct Instruction20 minutes
Part 1: The parallel criterion. Two distinct non-vertical lines are parallel if and only if they have equal slopes. Prove both directions with algebra:
Equal slopes means parallel. Take y = mx + b and y = mx + c with b ≠ c. A common point would need mx + b = mx + c, so b = c, which is false. The lines share no point, so they are parallel.
Parallel means equal slopes. Suppose the slopes are different, m₁ ≠ m₂. Then m₁x + b₁ = m₂x + b₂ has the solution x = (b₂ - b₁)/(m₁ - m₂), so the lines meet. Parallel lines never meet, so their slopes cannot be different.
Vertical lines. Two distinct vertical lines x = h and x = k are parallel; they have no slope, so the criterion is stated for non-vertical lines.
Part 2: The perpendicular criterion. Two non-vertical lines are perpendicular if and only if the product of their slopes is -1. Translating a pair of lines does not change their slopes or the angle between them, so it is enough to prove the claim for lines through the origin, y = m₁x and y = m₂x. Use Diagram 1: the points P(1, m₁) and Q(1, m₂) lie on the two lines. By the Pythagorean Theorem and its converse, angle POQ is a right angle exactly when OP² + OQ² = PQ², that is, (1 + m₁²) + (1 + m₂²) = (m₁ - m₂)². Expanding the right side gives m₁² - 2m₁m₂ + m₂², and the equation simplifies to 2 = -2m₁m₂, so m₁m₂ = -1. Because every step is reversible, the argument proves both directions. A horizontal line and a vertical line are also perpendicular, but the vertical one has no slope, so this case is stated separately.
Part 3: Using the criteria. To write a line parallel or perpendicular to a given line through a given point: rewrite the given line in slope-intercept form to read its slope, take the same slope (parallel) or the negative reciprocal (perpendicular), substitute the point to find the intercept, and check that the point satisfies your equation. Diagram 2 shows both constructions for one line and one point.
Proving two lines parallel
Prove that y = 3x - 2 and y = 3x + 5 are parallel.
Equation: A common point needs 3x - 2 = 3x + 5, so -2 = 5, which is false. No common point: the lines are parallel.
Classifying a pair of lines
Are 4x - 6y = 7 and 3x + 2y = 1 parallel, perpendicular or neither?
Equation: Slopes 2/3 and -3/2; (2/3)(-3/2) = -1, so the lines are perpendicular.
Parallel line through a point
Find the line parallel to 5x + 2y = 8 that passes through (4, -1).
Equation: Slope -5/2; -1 = (-5/2)(4) + b gives b = 9, so y = -5x/2 + 9.
Perpendicular line through a point
Find the line perpendicular to y = -3x + 4 that passes through (6, 1).
Equation: Slope 1/3; 1 = (1/3)(6) + b gives b = -1, so y = x/3 - 1.
Geometric problem: an altitude
Triangle ABC has A(0, 0), B(8, 4), C(2, 9). Find the equation of the altitude from C.
Equation: AB has slope 1/2, so the altitude has slope -2: y = -2x + 13. It meets AB at (5.2, 2.6).
Guided Practice15-20 minutes
Pairs complete four tasks. After each one, a pair explains its reasoning at the board.
Proof check: Use P(1, 5) and Q(1, -1/5) to verify with the Pythagorean Theorem that y = 5x and y = -x/5 are perpendicular. (OP² = 26, OQ² = 26/25, PQ² = (26/5)² = 676/25, and 26 + 26/25 = 676/25.)
Parallel: Find the line parallel to y = -4x + 3 through (-1, 9). (y = -4x + 5.)
Perpendicular: Find the line perpendicular to 2x + 5y = 10 through (-4, 3). (The given slope is -2/5, so the new slope is 5/2: y = 5x/2 + 13.)
Special case: Find the line perpendicular to y = 6 through (2, -3). (y = 6 is horizontal, so the answer is the vertical line x = 2.)
Listen for these errors: reading the slope of 2x + 5y = 10 as 2 without solving for y, taking the reciprocal without changing the sign, and using the given point's y-coordinate as the intercept.
Independent Practice15 minutes
Students work alone on three problems:
Classify each pair as parallel, perpendicular or neither: (a) y = 7x - 1 and 14x - 2y = 9 (parallel: both slopes 7, different intercepts), (b) 3x + 4y = 12 and 8x - 6y = 5 (perpendicular: -3/4 and 4/3), (c) y = 2x + 1 and y = -2x + 1 (neither: the product is -4).
Find the line parallel to 3x - y = 4 through (2, -5). (y = 3x - 11.)
Find the line perpendicular to y = 2x/3 + 1 through (4, -1). (y = -3x/2 + 5.)
Early finishers write, in their own words, why the proof of the perpendicular criterion is allowed to use only lines through the origin.
Closure5-10 minutes
Exit ticket: (1) In the Pythagorean proof, which equation turns into m₁m₂ = -1, and why does it describe a right angle? (OP² + OQ² = PQ²; by the converse of the Pythagorean Theorem it holds exactly when angle POQ is right.) (2) Write the line perpendicular to y = -x/4 + 2 through the origin. (y = 4x.) (3) Explain why the criterion m₁m₂ = -1 cannot be used for the lines x = 1 and y = 3, and say whether they are perpendicular.
Differentiation Strategies
For Struggling Students
Give a slope-finding checklist: solve for y first, then read the coefficient of x, including its sign
Use patty paper to rotate a drawn slope triangle 90° and count the new run and rise on the grid before writing any algebra
Provide a template for writing a line: slope = ___, point = (___, ___), substitute: ___ = ___ · ___ + b, so b = ___
For Advanced Students
Prove the perpendicular criterion with the dot product: the direction vectors (1, m₁) and (1, m₂) are perpendicular when 1 + m₁m₂ = 0
Find the distance from the point (7, 1) to the line y = 2x + 2 by finding the foot of the perpendicular, and compare with the formula |Ax₀ + By₀ + C|/√(A² + B²)
Show that the lines ax + by = c and bx - ay = d are always perpendicular, including when a or b is 0
Assessment Guidance
What to Look For
For the proofs, check that students argue both directions of each "if and only if" and that they explain why lines through the origin are enough (translations keep slopes and angles). A student who only checks one numerical pair has not proved the criterion. For equations, look for the slope read correctly from standard form, the negative reciprocal (not only the reciprocal), a check that the given point satisfies the final equation, and correct handling of horizontal and vertical lines.
02
Classroom Activities
3 Activities
1
Rotate the Slope Triangle
20 minPairs
Students give a second proof of the perpendicular criterion with a rotation. They draw a line and its slope triangle, trace them on patty paper, rotate the tracing 90° about a point on the line, and read the new slope from the grid.
Procedure
Draw a line with slope 3/2 through the origin and a slope triangle with run 2 and rise 3
Trace the line and triangle, then rotate the tracing 90° counterclockwise about the origin. The run of 2 becomes a rise of 2, and the rise of 3 becomes a run of -3, so the new slope is 2/(-3) = -2/3
Repeat with slopes 7 (new slope -1/7) and -1/3 (new slope 3), and record each product of slopes: it is -1 every time
Generalize: a slope triangle with run a and rise b (a, b ≠ 0) rotates to run -b and rise a, so the slopes are b/a and a/(-b), and their product is -1
Discussion Questions
Why does a 90° rotation give a line perpendicular to the original one?
Where in the general argument do we need b ≠ 0? What line do we get when b = 0?
How could a translation of a slope triangle explain why parallel lines have equal slopes?
Modification for Distance Learning
Use dynamic geometry software: students construct a line, its slope triangle and its image under a 90° rotation, then drag the line and watch the product of the two slopes stay at -1.
2
Line Builder Relay
20 minGroups of 4
Each group receives 8 cards. Each card gives a line and a point and asks for the parallel or perpendicular line through the point. Students rotate roles on every card: one finds the slope, one finds the intercept, one writes the equation, and one checks that the point satisfies it.
The 8 Cards (with answers for the teacher)
Card 1: parallel to y = 2x - 7 through (3, 1). (y = 2x - 5)
Card 2: perpendicular to y = 2x - 7 through (4, 0). (y = -x/2 + 2)
Card 3: parallel to x + y = 6 through (-2, 5). (y = -x + 3)
Card 4: perpendicular to x + y = 6 through (1, 4). (y = x + 3)
Card 5: parallel to 3x + 4y = 8 through (4, -2). (y = -3x/4 + 1)
Card 6: perpendicular to 3x + 4y = 8 through (3, 5). (y = 4x/3 + 1)
Card 7: parallel to x = -5 through (2, 7). (x = 2)
Card 8: perpendicular to y = -1 through (-6, 4). (x = -6)
Procedure
Groups graph each given line and the new line on one grid to confirm the relationship visually
The checker must substitute the point into the final equation before the group moves on
The first group to finish all 8 correctly writes one new card for another group
Challenge Variation
Give each group a line in the form ax + by = c with letters and a point (h, k), and ask for the parallel and perpendicular lines in the same form. (Parallel: ax + by = ah + bk. Perpendicular: bx - ay = bh - ak.)
3
Geometry Detective
15 minGroups of 3
Groups solve three short geometric problems where the slope criteria are the key tool. Each group writes a one-sentence justification that names the criterion it used.
The Three Cases (with answers for the teacher)
Case 1: Classify the quadrilateral A(-3, -1), B(3, 1), C(2, 4), D(-1, 3). (AB and DC both have slope 1/3, so they are parallel; BC has slope -3 and AD has slope 2, so they are not. Since (1/3)(-3) = -1, AB and DC are both perpendicular to BC: ABCD is a right trapezoid.)
Case 2: Find the perpendicular bisector of the segment from (-2, 3) to (6, -1). (Midpoint (2, 1), segment slope -1/2, bisector slope 2: y = 2x - 3.)
Case 3: Find k so that the line through (1, k) and (4, 7) is perpendicular to y = 3x - 2. (The slope must be -1/3, so (7 - k)/3 = -1/3 and k = 8.)
Procedure
Groups have 5 minutes per case and must sketch each figure on graph paper
After each case, one group reads its justification aloud, and the class checks that it names the parallel or perpendicular criterion
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Proving the Perpendicular Slope Criterion
The lines y = 2x and y = -x/2 pass through the origin. The points P(1, 2) and Q(1, -1/2) lie on them, directly above and below (1, 0). Because OP² + OQ² = PQ², the converse of the Pythagorean Theorem shows that angle POQ is a right angle. The same computation with P(1, m₁) and Q(1, m₂) proves that two lines are perpendicular exactly when m₁m₂ = -1. Drawn to scale.
Diagram 2: A Parallel and a Perpendicular Line Through a Point
The given line y = x/2 + 1 and the point P(2, 6). The parallel line keeps the slope 1/2, and the perpendicular line has slope -2, the negative reciprocal. Substituting P gives y = x/2 + 5 and y = -2x + 10. The perpendicular meets the given line at F(3.6, 2.8), the point of the line closest to P. Drawn to scale.
04
Homework Assignment
~30 min
HSG.GPE.B.5 Homework: Slope Criteria for Parallel and Perpendicular Lines
Directions: Show all work. In Part 1, write each proof in complete sentences. In Parts 2 and 3, state the slope you use and why, and check that every given point satisfies the equation you write. Give equations in slope-intercept form unless the line is vertical.
Part 1: Proving the Criteria (Problems 1-2)
Prove that if two lines y = mx + b and y = mx + c have the same slope and b ≠ c, they never intersect. Then use your argument to show that y = -2x + 7 and 6x + 3y = 4 are parallel.
The lines y = m₁x and y = m₂x pass through P(1, m₁) and Q(1, m₂). Show that angle POQ is a right angle exactly when m₁m₂ = -1, using the Pythagorean Theorem and its converse. Then test your result with m₁ = 3 and m₂ = -1/3 by computing OP², OQ² and PQ².
Part 2: Writing Equations (Problems 3-4)
Find the equation of the line through (10, -3) that is (a) parallel to 4x - 5y = 20 and (b) perpendicular to 4x - 5y = 20.
(a) Find the equation of the line through (1, 1) that is perpendicular to the line through (-3, 2) and (5, -2). (b) Find the equation of the line through (3, 8) that is perpendicular to x = -4.
Part 3: Geometric Problems (Problems 5-6)
Three vertices of parallelogram ABCD are A(-2, -1), B(4, 1) and C(5, 5). Find the coordinates of D, and use slopes to prove that ABCD is a parallelogram. Is it a rectangle? Justify your answer.
Find the equation of the perpendicular bisector of the segment with endpoints A(-1, 7) and B(5, -1). Then show that the point (6, 6) lies on the bisector and is the same distance from A and B.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Proofs (Part 1)
Both arguments complete and clearly reasoned, numerical checks correct
Correct idea with a gap or an algebra slip
Only numerical examples, or no proof
Slopes
Every slope read or computed correctly, negative reciprocals used for perpendicular lines
One slope error
Several slope errors
Equations
All equations correct and checked with the given point
Correct method with an intercept error
Missing or incorrect
Geometric Reasoning (Part 3)
Vertex, bisector and justifications correct and explained
Correct results without justification
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
What is the slope of a line parallel to 6x + 2y = 5?
Answer: B
Solve for y: 2y = -6x + 5, so y = -3x + 5/2 and the slope is -3. A parallel line has the same slope, -3. Choice A loses the sign when moving 6x across. Choice C is the slope of a perpendicular line. Choice D reads the coefficient of x without solving for y.
Question 2 of 20 · Multiple Choice
What is the slope of a line perpendicular to y = -(2/7)x + 1?
Answer: A
The negative reciprocal of -2/7 is 7/2, and (-2/7)(7/2) = -1. Choice B is the reciprocal without the sign change; its product with -2/7 is +1. Choice D is the slope of a parallel line.
Question 3 of 20 · Multiple Choice
Which pair of lines is perpendicular?
Answer: D
The slopes 4 and -1/4 have product -1. Choice A is a pair of parallel lines. Choice B only changes the sign (product -16), and choice C only takes the reciprocal (product 1).
Question 4 of 20 · Multiple Choice
Which line passes through (3, -2) and is parallel to y = 5x + 1?
Answer: A
The parallel line has slope 5. Substitute (3, -2): -2 = 5(3) + b, so b = -17. Choice B comes from -2 + 15 instead of -2 - 15. Choice C is the perpendicular line through the point. Choice D uses the y-coordinate of the point as the intercept.
Question 5 of 20 · Multiple Choice
Which line passes through (-3, 2) and is perpendicular to y = (3/4)x - 2?
Answer: B
The perpendicular slope is -4/3. Substitute (-3, 2): 2 = (-4/3)(-3) + b = 4 + b, so b = -2. Choice A uses the reciprocal without the sign change. Choice C is the parallel line through the point. Choice D has the right slope, but -(4/3)(-3) + 6 = 10, so it misses the point.
Question 6 of 20 · Multiple Choice
In a proof that y = 4x - 1 and y = 4x + 6 are parallel, what happens when you set 4x - 1 equal to 4x + 6?
Answer: C
Subtracting 4x from both sides leaves -1 = 6, which is false. So the system has no solution, the lines have no common point, and distinct lines in a plane with no common point are parallel. Choice B moves 4x to the left side without changing its sign. Choice D would happen only if the two intercepts were equal, when the equations describe the same line.
Question 7 of 20 · Multiple Choice
In the proof of the perpendicular criterion, the lines y = m₁x and y = m₂x pass through P(2, 2m₁) and Q(2, 2m₂). Which equation says that angle POQ is a right angle?
Answer: C
OP² = 2² + (2m₁)² = 4 + 4m₁², OQ² = 4 + 4m₂² and PQ² = (2m₁ - 2m₂)², since P and Q are on the vertical line x = 2. By the Pythagorean Theorem and its converse, the angle at O is right exactly when OP² + OQ² = PQ², which simplifies to 8 = -8m₁m₂, so m₁m₂ = -1. Choice A forgets to square. Choice D multiplies the squared lengths instead of adding them.
Question 8 of 20 · Multiple Choice
A slope triangle for a line has a run of 3 and a rise of 2. After a 90° counterclockwise rotation about a point on the line, what are the run and rise of the image triangle?
Answer: C
A 90° counterclockwise rotation sends (3, 2) to (-2, 3), so the new run is -2 and the rise is 3. The new slope is 3/(-2) = -3/2, and (2/3)(-3/2) = -1. Choice A swaps without the sign change (that is a reflection over y = x). Choice B is a 180° rotation, which gives the same slope.
Question 9 of 20 · Multiple Choice
How are the lines 2x - 3y = 6 and 3x + 2y = 4 related?
Answer: D
Solving for y gives slopes 2/3 and -3/2. Their product is -1, so the lines are perpendicular. A common error is reading the slopes as 2 and 3 straight from the coefficients; always solve for y first.
Question 10 of 20 · Multiple Choice
For which value of k is the line y = kx + 2 perpendicular to 8x - y = 1?
Answer: D
The line 8x - y = 1 is y = 8x - 1, with slope 8. Perpendicular means 8k = -1, so k = -1/8. Choice A gives a parallel line. Choice B changes only the sign and choice C only takes the reciprocal.
Question 11 of 20 · Multiple Choice
Which line passes through (-1, 4) and is perpendicular to x = 3?
Answer: B
x = 3 is a vertical line, so every line perpendicular to it is horizontal, and the horizontal line through (-1, 4) is y = 4. Choice A is vertical, so it is parallel to x = 3. Choice C treats 3 as a slope, but x = 3 has no slope.
Question 12 of 20 · Multiple Choice
Quadrilateral PQRS has vertices P(0, 0), Q(5, 1), R(6, 4) and S(1, 3). Which description do the slopes prove?
Answer: A
PQ and RS both have slope 1/5, and QR and SP both have slope 3, so both pairs of opposite sides are parallel: a parallelogram. The product (1/5)(3) = 3/5 is not -1, so adjacent sides are not perpendicular and it is not a rectangle. Choice C ignores the second pair of parallel sides.
Question 13 of 20 · Multiple Choice
A student says that the line perpendicular to y = (2/5)x + 1 has slope 5/2. What is the error?
Answer: B
Perpendicular slopes must multiply to -1, and (2/5)(5/2) = 1, not -1. The correct slope is -5/2, since (2/5)(-5/2) = -1. Choice C changes the sign without taking the reciprocal. Choice D describes parallel lines.
Question 14 of 20 · Multiple Choice
Which statement about the lines y = -3 and x = 2 is true?
Answer: A
y = -3 is horizontal and x = 2 is vertical, so they meet at (2, -3) at a right angle. The slope criterion is stated for non-vertical lines, so this case is handled separately. Choice C treats the vertical line as if it had slope 0. Choice D is false: they intersect at (2, -3).
Question 15 of 20 · Short Answer
Prove that the lines y = x/2 + 3 and x - 2y = 8 are parallel.
Solve the second equation for y: y = x/2 - 4. Both lines have slope 1/2 and different intercepts. A common point would need x/2 + 3 = x/2 - 4, so 3 = -4, which is false. The lines have no common point, so they are parallel.
Question 16 of 20 · Short Answer
The lines y = (2/3)x and y = -(3/2)x pass through P(1, 2/3) and Q(1, -3/2). Compute OP², OQ² and PQ², and use them to show that the lines are perpendicular.
OP² = 1 + 4/9 = 13/9. OQ² = 1 + 9/4 = 13/4. PQ² = (2/3 + 3/2)² = (13/6)² = 169/36. Then OP² + OQ² = 52/36 + 117/36 = 169/36 = PQ². By the converse of the Pythagorean Theorem, angle POQ is a right angle, so the lines are perpendicular. This matches (2/3)(-3/2) = -1.
Question 17 of 20 · Short Answer
Find the equation of the line through (-3, -5) that is perpendicular to 3x + y = 9.
The given line is y = -3x + 9, with slope -3, so the perpendicular slope is 1/3. Substitute (-3, -5): -5 = (1/3)(-3) + b = -1 + b, so b = -4. y = x/3 - 4. Check: (1/3)(-3) - 4 = -5.
Question 18 of 20 · Short Answer
Triangle ABC has vertices A(1, 1), B(7, 4) and C(3, -3). Use slopes to prove that it is a right triangle, and name the right angle.
The slope of AB is (4 - 1)/(7 - 1) = 1/2, and the slope of AC is (-3 - 1)/(3 - 1) = -2. Since (1/2)(-2) = -1, AB is perpendicular to AC. Triangle ABC is a right triangle with the right angle at A.
Question 19 of 20 · Short Answer
Find the equation of the line through (4, 0) that is parallel to the line through (-1, 2) and (3, 8).
The slope of the line through (-1, 2) and (3, 8) is 6/4 = 3/2. The parallel line through (4, 0): 0 = (3/2)(4) + b, so b = -6. y = 3x/2 - 6.
Question 20 of 20 · Short Answer
For which value of a is the line ax + 3y = 7 parallel to y = -2x + 1? Check that the two lines are not the same line.
Solving ax + 3y = 7 for y gives y = -(a/3)x + 7/3, with slope -a/3. Parallel means -a/3 = -2, so a = 6. The lines are then y = -2x + 7/3 and y = -2x + 1, which have different intercepts, so they are distinct parallel lines.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GPE.B.5 mean?
It means students must prove why parallel lines have equal slopes and why perpendicular lines have slopes whose product is -1, and then use these facts to solve problems. The typical problem named in the standard is writing the equation of a line parallel or perpendicular to a given line through a given point.
Is HSG.GPE.B.5 taught in Geometry or Algebra 1?
HSG.GPE.B.5 is a Geometry standard and is usually taught in Geometry, often in a unit on coordinate geometry. Many Algebra I courses already use the slope rules for parallel and perpendicular lines, but the proof of the rules and their use in geometric problems belong to this standard.
How do you prove that parallel lines have equal slopes?
Use algebra on two lines y = m₁x + b₁ and y = m₂x + b₂. If m₁ ≠ m₂, the equation m₁x + b₁ = m₂x + b₂ has the solution x = (b₂ - b₁)/(m₁ - m₂), so the lines meet and are not parallel. If m₁ = m₂ and b₁ ≠ b₂, the equation becomes b₁ = b₂, which is false, so the lines never meet. A second argument uses slope triangles: a translation moves one slope triangle onto the other, so the triangles are congruent and the slopes are equal.
How do you prove the perpendicular slope criterion?
Move the lines so they meet at the origin (a translation keeps slopes and angles), and write them as y = m₁x and y = m₂x. The points P(1, m₁) and Q(1, m₂) are on the lines. By the Pythagorean Theorem and its converse, the angle at the origin is right exactly when (1 + m₁²) + (1 + m₂²) = (m₁ - m₂)², which simplifies to m₁m₂ = -1. A rotation argument also works: a 90° rotation turns a slope triangle with run a and rise b into one with run -b and rise a.
What is a negative reciprocal?
The negative reciprocal of a nonzero number m is -1/m: flip the fraction and change its sign. For example, the negative reciprocal of 3/4 is -4/3, and that of -5 is 1/5. A number times its negative reciprocal is always -1, which is exactly the perpendicular criterion.
What about horizontal and vertical lines?
A horizontal line has slope 0 and a vertical line has no slope, so the product rule cannot be applied to them. They are still perpendicular to each other. Every line perpendicular to a horizontal line y = k is vertical (x = h), and every line perpendicular to a vertical line is horizontal. Two vertical lines are parallel even though neither has a slope.
How do you find the equation of a line perpendicular to a given line through a point?
Find the slope of the given line (solve for y if needed), take its negative reciprocal, and substitute the point into y = mx + b to find b. For example, the line perpendicular to 2x - y = 3 (slope 2) through (4, 1) has slope -1/2, and 1 = (-1/2)(4) + b gives b = 3, so y = -x/2 + 3. Always check that the point satisfies the final equation.
What are common mistakes with parallel and perpendicular slopes?
A common mistake is taking the reciprocal without changing the sign, so that 4/9 becomes 9/4 instead of -9/4. Others are:
reading the slope of 3x + 4y = 8 as 3 instead of solving for y (slope -3/4)
using the y-coordinate of the given point as the y-intercept
trying to use m₁m₂ = -1 with a vertical line
checking one example and calling it a proof of the criterion
Where are the slope criteria used in geometry problems?
They are the main tool in coordinate proofs (HSG.GPE.B.4): showing that a quadrilateral has parallel sides or right angles, finding altitudes of triangles, writing perpendicular bisectors, finding a missing vertex of a parallelogram or rectangle, and finding the point of a line closest to a given point. They are also used when computing areas with coordinates (HSG.GPE.B.7), where the height of a triangle lies on a perpendicular.
How does HSG.GPE.B.5 connect to earlier and later courses?
It builds on grade 8, where students explain slope with similar triangles (8.EE.B.6) and prove the Pythagorean Theorem and its converse (8.G.B.6). Later, the same ideas appear in vectors, where perpendicular directions have a dot product of 0, and in calculus, where the normal line to a curve is perpendicular to the tangent line.
07
Related Standards
5 standards
These standards connect to HSG.GPE.B.5: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.EE.B.6Prerequisite
Use similar triangles to explain why slope is the same between any two points on a line
