HSN.CN.B.6Common CoreMathNumber and QuantityGrades 9-12
HSN.CN.B.6: Distance and Midpoint in the Complex Plane
In plain English: HSN.CN.B.6 is an advanced (+) Common Core number and quantity standard, usually taught in Precalculus. Students find the distance between two complex numbers z and w as the modulus of their difference, |z - w|, and the midpoint of the segment joining them as their average, (z + w)/2, and they explain both rules with the geometry of the complex plane.
(+) Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints.
Common Core State Standards for Mathematics · Domain: The Complex Number System (CN) · Cluster: Represent complex numbers and their operations on the complex plane. Also written as HSN-CN.B.6 or N-CN.6 · Official standard
Students already know how to find the distance and the midpoint between two points in the coordinate plane. In this lesson they learn to do the same work with complex numbers as single objects. The distance between z and w is |z - w|: subtracting slides the segment from w to z so that it starts at 0, and the modulus measures its length. The midpoint is (z + w)/2: the average of the two numbers, which averages the real parts and the imaginary parts at the same time.
Students justify both rules with pictures, the parallelogram whose diagonals bisect each other and the arrow z - w, and then use them to solve map problems, find missing endpoints, and classify triangles.
Learning Objectives
By the end of this lesson, students will be able to:
Calculate the distance between two complex numbers z and w as |z - w|
Explain why |z - w| equals the length of the segment from w to z and why |z - w| = |w - z|
Calculate the midpoint of the segment joining z and w as (z + w)/2 and explain why the average lands halfway
Use distance and midpoint to solve problems, such as finding a missing endpoint or showing that a triangle is isosceles
Prior Knowledge Required
Students should already be comfortable with:
Using the Pythagorean Theorem to find the distance between two points 8.G.B.8
Start from what students know in the coordinate plane.
Warm-Up Prompt
"Find the distance between (1, 2) and (7, 10), and find the point halfway between them. Now rename the points as the complex numbers 1 + 2i and 7 + 10i. Can you get the same two answers with one subtraction and one addition?"
Students should get a distance of √(6² + 8²) = 10 and the midpoint (4, 6). Some students will notice that (7 + 10i) - (1 + 2i) = 6 + 8i contains the two legs of the right triangle, and that (1 + 2i) + (7 + 10i) = 8 + 12i is twice the midpoint. Collect both observations; they are the two rules of the lesson.
Direct Instruction20 minutes
Part 1: Distance as the modulus of the difference. Use Diagram 1. For z = a + bi and w = c + di, the difference is z - w = (a - c) + (b - d)i. Its real part is the horizontal change from w to z and its imaginary part is the vertical change, so
Subtract: find z - w. The order does not matter for distance, because w - z = -(z - w) points the opposite way and has the same length.
Take the modulus: |z - w| = √((a - c)² + (b - d)²), which is exactly the distance formula.
Interpret: z - w is the arrow from w to z, slid so that it starts at 0. Sliding does not change length, and the modulus is the distance from 0, so |z - w| is the length of the segment from w to z.
Midpoint as an average: the midpoint is m = (z + w)/2 = (a + c)/2 + ((b + d)/2)i, averaging the real parts and the imaginary parts.
Check the midpoint: m - w = (z - w)/2, so m is half of the way from w to z, and |z - m| = |m - w| = |z - w|/2.
Part 2: Why the average is the midpoint. Use Diagram 2. The points 0, z, z + w and w form a parallelogram, and the diagonals of a parallelogram bisect each other. The diagonal from 0 to z + w has midpoint (z + w)/2, so the other diagonal, the segment from z to w, has the same midpoint. Then work the examples below.
Distance
Find the distance between z = 5 + 3i and w = 1 + 6i.
Equation: z - w = 4 - 3i, so |z - w| = √(16 + 9) = 5
Distance with negative parts
Find the distance between -2 - 3i and 4 + 5i. Subtract in either order.
Equation: (1 + 2i) - (4 - 4i) = -3 + 6i, so the distance is √45 = 3√5 ≈ 6.71
Distance and midpoint together
A circle has a diameter with endpoints -4 + i and 2 + 9i. Find its center and radius.
Equation: Center (-4 + i + 2 + 9i)/2 = -1 + 5i; radius |6 + 8i|/2 = 5
Guided Practice15 minutes
Pairs compute each value and mark it on a sketch. One partner subtracts in the order z - w and the other in the order w - z, then they compare.
Guided practice
Numbers
Find
Result
6 - 2i and 1 + 10i
Distance
|5 - 12i| = 13
6 - 2i and 1 + 10i
Midpoint
3.5 + 4i
3 + 4i and its conjugate
Distance
|8i| = 8
2i and -7
Distance
|7 + 2i| = √53 ≈ 7.28
Watch for students who add the numbers when they should subtract, or who drop the square root and report 169 for the first distance. For the conjugate pair, ask why the distance is always twice the absolute value of the imaginary part: the two points are mirror images across the real axis.
Independent Practice10-15 minutes
Students work alone: (1) the distance between 9 + i and 1 - 5i (10) and (2) their midpoint (5 - 2i); (3) the distance from 0 to -5 + 12i (13), noting that the modulus is a distance from 0; (4) the midpoint of -6 + 3i and 2 - 7i (-2 - 2i); (5) the distance between -1 - i and 2 + 3i (5); (6) the midpoint of 4i and 10 (5 + 2i). As a stretch item, find the distance between 3 + 3i and -3 - 3i (6√2).
Closure5 minutes
Exit ticket: (1) Find the distance between 7 + 2i and 4 - 2i. (Answer: |3 + 4i| = 5.) (2) Find their midpoint. (Answer: 5.5, a real number.) (3) In one sentence, explain why |z - w| and |w - z| are always equal.
Differentiation Strategies
For Struggling Students
Have students draw the right triangle for each distance and label the legs with the real and imaginary parts of z - w
Use a two-column organizer: real parts in one column, imaginary parts in the other, for both subtracting and averaging
Start with pairs of numbers that share a real part or an imaginary part, so the distance can be counted on the grid
For Advanced Students
Ask students to describe the set of all z with |z - (1 + i)| = |z - (5 + 3i)|, and to explain why it is the perpendicular bisector of a segment
Ask students to show that the point (2z + w)/3 is one third of the way from z to w, using distances
Ask students to prove, with complex numbers, that the midpoints of the sides of any quadrilateral form a parallelogram
Assessment Guidance
What to Look For
Look for the difference inside the modulus, not the sum, and for the square root at the end. Students should give exact answers such as 3√5 before rounding. For midpoints, check that both parts are divided by 2. Ask each student to explain one rule with a picture: the arrow z - w for distance, or the parallelogram diagonals for the midpoint.
02
Classroom Activities
3 Activities
1
Town Map in the Complex Plane
20 minGroups of 3-4
Groups use a town map drawn on a complex plane grid, where 1 unit is 1 kilometer, to find straight-line distances and meeting points between four places.
The Map
Library L = 2 + 3i
School S = 8 - 5i
Park P = -4 + i
Train station T = -1 - 3i
Procedure
Each group member computes the straight-line distance for some of the six pairs, using |z - w|. Answers: L to S is 10 km, P to T is 5 km, L to P is 2√10 ≈ 6.32 km, S to T is √85 ≈ 9.22 km, L to T is 3√5 ≈ 6.71 km, and P to S is 6√5 ≈ 13.42 km
The group checks one distance with a ruler on the printed map
Find the halfway meeting point for L and S (5 - i) and for P and T (-2.5 - i), and mark both on the map
Discussion Questions
Which two places are closest? Which are farthest?
Real roads are not straight. Is the straight-line distance longer or shorter than a walking route, and why?
2
Find the Missing Endpoint
15 minPairs
Pairs get 4 cards. Each card gives a midpoint m and one endpoint z, and the pair must find the other endpoint w. Since m = (z + w)/2, the missing endpoint is w = 2m - z.
The 4 Cards
m = 1 + i, z = 4 - 3i (w = -2 + 5i)
m = -2 + 3i, z = 0 (w = -4 + 6i)
m = 3.5 - 0.5i, z = 2 + 4i (w = 5 - 5i)
m = -i, z = -6 + 2i (w = 6 - 4i)
Procedure
Solve for w, then check by averaging z and w
Check with distances: |z - m| and |w - m| must be equal
Sketch each card: the midpoint must sit on the segment, halfway
Modification for Distance Learning
Share the cards on a slide with a complex plane background. Students drag a point to where they think w is, then type the computation in a comment to confirm.
3
Classify a Triangle with Complex Numbers
15 minSmall groups
Groups use distances and midpoints to decide what kind of triangle three complex numbers form, and they present the argument on a poster.
Procedure
Plot A = 1 + i, B = 5 + 2i and C = 2 + 5i
Compute |B - A| = √17, |C - A| = √17 and |C - B| = √18. Conclude that the triangle is isosceles with apex A
Check whether it is a right triangle: 17 + 17 = 34, which is not 18, so it is not
Find the midpoint of BC, 3.5 + 3.5i, and the length of the median from A, |2.5 + 2.5i| = 2.5√2
Challenge Variation
Give the quadrilateral with vertices P = -3 + i, Q = 2 + 2i, R = 4 + 6i and S = -1 + 5i. Groups show it is a parallelogram by finding that the diagonals PR and QS have the same midpoint, 0.5 + 3.5i.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Distance as the Modulus of the Difference
The segment from w = 1 + 6i to z = 5 + 3i has horizontal change 4 and vertical change -3, the parts of z - w = 4 - 3i. The green arrow from 0 to 4 - 3i is the same segment slid to start at 0, so its length |z - w| = 5 is the distance from w to z. Drawn to scale, one grid square is one unit.
Diagram 2: The Midpoint as an Average
The points 0, z = -3 + 7i, z + w = 2 + 6i and w = 5 - i form a parallelogram. Its diagonals bisect each other, so the midpoint of the segment from z to w equals the midpoint of the diagonal from 0 to z + w, which is (z + w)/2 = 1 + 3i. Drawn to scale, one grid square is one unit.
04
Homework Assignment
~30 min
HSN.CN.B.6 Homework: Distance and Midpoint
Directions: Show the subtraction or addition you use in every problem, and sketch the points on a complex plane. Give exact answers first, then a decimal rounded to two places when the answer is not a whole number.
Part 1: Computing Distance and Midpoint (Problems 1-3)
Find the distance between each pair as the modulus of their difference: (a) 7 + 5i and 2 - 7i (b) -3 + 2i and 5 + 2i (c) -4i and 2 + i.
Find the midpoint of the segment joining each pair: (a) 8 - 3i and -2 + 9i (b) -5 - 4i and 1 + 7i (c) 6i and -9.
Let z = a + bi and w = c + di. Write z - w in the form x + yi and use the modulus to show that |z - w| is the distance formula. Then explain, using a sketch, why |z - w| = |w - z|.
Part 2: Applying the Rules (Problems 4-6)
A park map is drawn on a complex plane where 1 unit is 100 meters. A fountain is at 2 + 5i and a picnic area is at -6 - i. Find the straight-line distance between them in meters. A bench will be placed exactly halfway: give its location as a complex number and describe it in meters east or west and north or south of the origin.
The midpoint of a segment is 2 - 3i and one endpoint is -1 + 4i. Find the other endpoint, and check your answer by averaging.
A triangle has vertices A = -1 + 2i, B = 5 + 2i and C = 2 + 8i. Use distances to show that the triangle is isosceles. Then find the midpoint M of AB and the distance from C to M.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Distance
Modulus of the difference computed correctly and exactly
Correct setup with one arithmetic error
Sum used or modulus missing
Midpoint
Average computed with both parts halved
One part not halved or one sign error
Incorrect method
Explanation
Clear reason for |z - w| = |w - z| and for the distance formula link
Partial reasoning
No explanation
Applications
Map, missing endpoint and triangle answered with units and checks
Most parts correct
Mostly 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 distance between 3 + 8i and -6 - 4i?
Answer: C
(3 + 8i) - (-6 - 4i) = 9 + 12i, and |9 + 12i| = √(81 + 144) = √225 = 15. Choice A comes from adding the numbers instead of subtracting: |-3 + 4i| = 5. Choice B adds the legs 9 and 12 without the Pythagorean Theorem. Choice D subtracts 4 instead of -4 in the imaginary part.
Question 2 of 20 · Multiple Choice
Which expression gives the distance between the complex numbers z and w?
Answer: D
The difference z - w is the segment from w to z moved to start at 0, and its modulus is its length. Choice A compares distances from 0, which can be equal for points far apart (such as 1 and -1). Choice C is the midpoint, not a distance.
Question 3 of 20 · Multiple Choice
What is the midpoint of the segment joining 7 - 2i and -3 + 8i?
Answer: A
Average the numbers: ((7 - 2i) + (-3 + 8i))/2 = (4 + 6i)/2 = 2 + 3i. Choice C is the sum, not divided by 2. Choice B is half of the difference (7 - 2i) - (-3 + 8i), which is not a point on the segment.
Question 4 of 20 · Multiple Choice
Which expression gives the midpoint of the segment joining z and w?
Answer: C
The midpoint is the average of the endpoints, (z + w)/2. Choice A is half of the arrow from w to z, which starts at 0 rather than on the segment. Choice B divides only w by 2. Choice D is a real number, a length, not a point.
Question 5 of 20 · Multiple Choice
What is the distance between -2 + 5i and 4 + 5i?
Answer: B
(4 + 5i) - (-2 + 5i) = 6, and |6| = 6. The two points have the same imaginary part, so the segment is horizontal and 6 units long. Choice A subtracts 4 - 2 and ignores the sign of -2. Choice C is not a distance: distances are nonnegative real numbers.
Question 6 of 20 · Multiple Choice
What is the distance between -3 - 2i and 1 + 6i?
Answer: A
(1 + 6i) - (-3 - 2i) = 4 + 8i, and |4 + 8i| = √(16 + 64) = √80 = 4√5 ≈ 8.94. Choice B adds the legs. Choice C is half the distance, which is the distance from an endpoint to the midpoint. Choice D simplifies √80 incorrectly.
Question 7 of 20 · Multiple Choice
How far is -8 + 15i from 0 on the complex plane?
Answer: C
The distance from 0 is |(-8 + 15i) - 0| = √(64 + 225) = √289 = 17. Choice A adds -8 and 15. Choice B adds 8 and 15. Choice D subtracts the squares, √(225 - 64).
Question 8 of 20 · Multiple Choice
What is the midpoint of -4 + i and 9 + 6i?
Answer: B
((-4 + i) + (9 + 6i))/2 = (5 + 7i)/2 = 2.5 + 3.5i. Choice A forgets to divide by 2. Choice C is half of the difference (9 + 6i) - (-4 + i), and choice D is half of the difference in the other order.
Question 9 of 20 · Multiple Choice
The midpoint of the segment from z = 3 + 2i to w is -1 + 4i. What is w?
Answer: D
From (z + w)/2 = m, w = 2m - z = (-2 + 8i) - (3 + 2i) = -5 + 6i. Check: ((3 + 2i) + (-5 + 6i))/2 = -1 + 4i. Choice A averages m and z, which gives a point between them. Choice B is 2m and forgets to subtract z. Choice C adds z instead of subtracting it.
Question 10 of 20 · Multiple Choice
A student finds the distance between 2 + 3i and 5 - i by computing |7 + 2i|. What is the correct distance?
Answer: A
The student added the numbers. The distance uses the difference: (2 + 3i) - (5 - i) = -3 + 4i, and |-3 + 4i| = 5. Choice B is the student's value |7 + 2i|. Choice D is (-3)² + 4² with the square root left off.
Question 11 of 20 · Multiple Choice
Why is |z - w| the distance between z and w?
Answer: B
Adding w to z - w gives z, so z - w is the translation that takes w to z. Translating a segment keeps its length, and the modulus of a number is its distance from 0. Choice A is false: z - w usually has an imaginary part. Choice C is false; for example |1 - (-1)| = 2 but |1| - |-1| = 0.
Question 12 of 20 · Multiple Choice
Which set of points is described by |z - (2 - i)| = 3?
Answer: C
|z - (2 - i)| is the distance from z to 2 - i, so the equation lists every point 3 units from 2 - i: a circle with center 2 - i and radius 3. Choice A reverses the signs of the center. Choice B squares the radius, which belongs in the equation (x - 2)² + (y + 1)² = 9.
Question 13 of 20 · Multiple Choice
For z = a + bi, what is the midpoint of the segment joining z and its conjugate z̄?
Answer: A
(z + z̄)/2 = ((a + bi) + (a - bi))/2 = 2a/2 = a. The conjugate is the reflection of z across the real axis, so the midpoint lies on the real axis directly below or above z. Choice C is true only when a = 0. Choice D is z itself.
Question 14 of 20 · Multiple Choice
A circle has a diameter with endpoints -1 + 4i and 7 - 2i. What are its center and radius?
Answer: D
The center is the midpoint ((-1 + 4i) + (7 - 2i))/2 = 3 + i. The diameter is |(7 - 2i) - (-1 + 4i)| = |8 - 6i| = 10, so the radius is 5. Choice A gives the diameter as the radius. Choice B is half of the difference, not the average. Choice C squares the radius.
Question 15 of 20 · Short Answer
Find the distance between 2 - 3i and -5 + 21i. Show the subtraction.
On a hiking map drawn on the complex plane, 1 unit is 1 kilometer. Ana is at -3 + 4i and Ben is at 9 - i. How far apart are they in a straight line, and where should they meet if each walks the same straight-line distance toward the other?
(9 - i) - (-3 + 4i) = 12 - 5i, and |12 - 5i| = √(144 + 25) = 13. They are 13 km apart. The halfway point is ((-3 + 4i) + (9 - i))/2 = 3 + 1.5i, 6.5 km from each hiker. On real trails the walking distance would be longer than the straight line.
Question 18 of 20 · Short Answer
Show that the triangle with vertices 0, 3 + i and 1 + 3i is isosceles. Which side is the base?
|3 + i - 0| = √10, |1 + 3i - 0| = √10, and |(3 + i) - (1 + 3i)| = |2 - 2i| = √8. Two sides have length √10, so the triangle is isosceles. The base is the side from 3 + i to 1 + 3i, with length √8 = 2√2.
Question 19 of 20 · Short Answer
The midpoint of a segment is 4 - i, and one endpoint is 7 + 3i. Find the other endpoint.
From (z + w)/2 = m, w = 2m - z = (8 - 2i) - (7 + 3i) = 1 - 5i. Check: ((7 + 3i) + (1 - 5i))/2 = (8 - 2i)/2 = 4 - i.
Question 20 of 20 · Short Answer
A circle has center -2 + 3i and passes through the point 4 - 5i. Find its radius, and find the other endpoint of the diameter that starts at 4 - 5i.
The radius is the distance from the center to the point: (4 - 5i) - (-2 + 3i) = 6 - 8i, and |6 - 8i| = √(36 + 64) = 10. The center is the midpoint of the diameter, so (4 - 5i + w)/2 = -2 + 3i and w = 2(-2 + 3i) - (4 - 5i) = -8 + 11i. Check: |(-8 + 11i) - (-2 + 3i)| = |-6 + 8i| = 10, also the radius.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.CN.B.6 mean?
It means students find distances and midpoints in the complex plane by working with the complex numbers directly. The distance between z and w is |z - w|, the modulus of their difference. The midpoint of the segment from z to w is (z + w)/2, the average of its endpoints.
Is HSN.CN.B.6 just the distance formula again?
The numbers come out the same, but the idea is new. Writing z - w and taking its modulus gives √((a - c)² + (b - d)²), which is the distance formula from 8.G.B.8 and Geometry. What HSN.CN.B.6 adds is the reason: subtraction is a translation and the modulus is a length, so distance is built from complex operations students already know.
Which course teaches HSN.CN.B.6, and what does the (+) mean?
It is usually taught in Precalculus, right after polar form and the geometry of complex operations. The (+) tells teachers that Common Core counts it as additional mathematics for students headed to advanced courses such as calculus, so not every high school student is expected to learn it.
Why is the distance between two complex numbers the modulus of their difference?
Because z - w is the arrow from w to z, slid so that it starts at 0. Sliding a segment does not change its length, and the modulus of a number is its distance from 0. So |z - w| is the length of the segment from w to z.
Does it matter whether I compute z - w or w - z?
No, not for distance. The two differences are opposites, so they point in opposite directions but have the same modulus. For example, (5 + i) - (2 + 5i) = 3 - 4i and (2 + 5i) - (5 + i) = -3 + 4i, and both have modulus 5. The order matters only when you need the direction from one point to the other.
Why is the midpoint the average of the two numbers?
Averaging finds the point halfway between. The picture is a parallelogram with vertices 0, z, z + w and w: its diagonals bisect each other, so the middle of the segment from z to w is the same point as the middle of the diagonal from 0 to z + w, which is (z + w)/2. Algebraically, the average halves the change from w to z.
How do you find a missing endpoint from the midpoint?
Solve m = (z + w)/2 for the unknown endpoint: w = 2m - z. For example, if the midpoint is 1 - 2i and one endpoint is 3 + i, the other endpoint is 2 - 4i - 3 - i = -1 - 5i. Check by averaging the two endpoints.
What does an equation like |z - c| = r describe?
It describes a circle with center c and radius r: every point z whose distance from c is r. This is where HSN.CN.B.6 meets the circle equations of HSG.GPE.A.1. For instance, |z - (1 + 2i)| = 4 is the circle (x - 1)² + (y - 2)² = 16.
What mistakes do students make with complex distance and midpoint?
A common one is using the sum z + w inside the modulus instead of the difference. Others include forgetting the square root in the modulus, dividing only the real part by 2 when averaging, and mixing up signs when subtracting a number with negative parts. A quick sketch shows most of these errors at once.
How does HSN.CN.B.6 connect to vectors and geometry?
The difference z - w works just like the vector from one point to another, found by subtracting coordinates (HSN.VM.A.2), and its modulus is the vector's magnitude. Distance and midpoint in complex form also give quick proofs in coordinate geometry, such as showing a triangle is isosceles or a quadrilateral is a parallelogram.
07
Related Standards
6 standards
These standards connect to HSN.CN.B.6: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.B.8Prerequisite
Use the Pythagorean Theorem to find the distance between two points
