Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  3. Algebra 2

  5. Flashcards

Algebra 2 Flashcards: Distance Midpoints In The Complex Plane

Study Distance Midpoints In The Complex Plane in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Distance Midpoints In The Complex Plane, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Algebra 2 Flashcards: Distance Midpoints In The Complex Plane

1

/ 30

0 reviewed

0% Complete

0 reviewing
QUESTION

Find the distance between z1=−2+5iz_1=-2+5iz1​=−2+5i and z2=−2−1iz_2=-2-1iz2​=−2−1i.

Tap or drag to reveal answer

ANSWER
  1. ∣(−2+5i)−(−2−1i)∣=∣6i∣=6|(-2+5i)-(-2-1i)| = |6i| = 6∣(−2+5i)−(−2−1i)∣=∣6i∣=6

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: Find the distance between z1=−2+5iz_1=-2+5iz1​=−2+5i and z2=−2−1iz_2=-2-1iz2​=−2−1i.

Answer:

  1. ∣(−2+5i)−(−2−1i)∣=∣6i∣=6|(-2+5i)-(-2-1i)| = |6i| = 6∣(−2+5i)−(−2−1i)∣=∣6i∣=6

Flashcard 2: Find the midpoint of z1=0z_1=0z1​=0 and z2=6+8iz_2=6+8iz2​=6+8i.

Answer: 3+4i3+4i3+4i. 0+(6+8i)2=6+8i2=3+4i\frac{0+(6+8i)}{2} = \frac{6+8i}{2} = 3+4i20+(6+8i)​=26+8i​=3+4i.

Flashcard 3: Find the distance between z1=0+2iz_1=0+2iz1​=0+2i and z2=0−4iz_2=0-4iz2​=0−4i.

Answer: 666. ∣(0+2i)−(0−4i)∣=∣6i∣=6|(0+2i)-(0-4i)| = |6i| = 6∣(0+2i)−(0−4i)∣=∣6i∣=6.

Flashcard 4: Find the midpoint of z1=−7+3iz_1=-7+3iz1​=−7+3i and z2=1−5iz_2=1-5iz2​=1−5i.

Answer: −3−i-3-i−3−i. (−7+3i)+(1−5i)2=−6−2i2=−3−i\frac{(-7+3i)+(1-5i)}{2} = \frac{-6-2i}{2} = -3-i2(−7+3i)+(1−5i)​=2−6−2i​=−3−i.

Flashcard 5: What is the distance between z1z_1z1​ and z2z_2z2​ if z1−z2=−8+6iz_1-z_2=-8+6iz1​−z2​=−8+6i?

Answer: 101010. ∣−8+6i∣=64+36=10|-8+6i| = \sqrt{64+36} = 10∣−8+6i∣=64+36​=10.

Flashcard 6: Find the distance between z1=2−4iz_1=2-4iz1​=2−4i and z2=−6+0iz_2=-6+0iz2​=−6+0i.

Answer: 80\sqrt{80}80​. ∣(2−4i)−(−6+0i)∣=∣8−4i∣=64+16=80|(2-4i)-(-6+0i)| = |8-4i| = \sqrt{64+16} = \sqrt{80}∣(2−4i)−(−6+0i)∣=∣8−4i∣=64+16​=80​.

Flashcard 7: Find the distance between z1=−3+0iz_1=-3+0iz1​=−3+0i and z2=5+0iz_2=5+0iz2​=5+0i.

Answer: 888. ∣(−3+0i)−(5+0i)∣=∣−8∣=8|(-3+0i)-(5+0i)| = |-8| = 8∣(−3+0i)−(5+0i)∣=∣−8∣=8.

Flashcard 8: Find ∣z∣|z|∣z∣ for z=9+12iz=9+12iz=9+12i.

Answer: 151515. ∣9+12i∣=81+144=225=15|9+12i| = \sqrt{81+144} = \sqrt{225} = 15∣9+12i∣=81+144​=225​=15.

Flashcard 9: Find ∣z∣|z|∣z∣ for z=5−12iz=5-12iz=5−12i.

Answer: 131313. ∣5−12i∣=25+144=169=13|5-12i| = \sqrt{25+144} = \sqrt{169} = 13∣5−12i∣=25+144​=169​=13.

Flashcard 10: Find the distance between z1=1+2iz_1=1+2iz1​=1+2i and z2=4+6iz_2=4+6iz2​=4+6i.

Answer: 555. ∣(1+2i)−(4+6i)∣=∣−3−4i∣=9+16=5|(1+2i)-(4+6i)| = |-3-4i| = \sqrt{9+16} = 5∣(1+2i)−(4+6i)∣=∣−3−4i∣=9+16​=5.

Flashcard 11: Find the distance between z1=−2+1iz_1=-2+1iz1​=−2+1i and z2=4−7iz_2=4-7iz2​=4−7i.

Answer: 101010. ∣(−2+1i)−(4−7i)∣=∣−6+8i∣=36+64=10|(-2+1i)-(4-7i)| = |-6+8i| = \sqrt{36+64} = 10∣(−2+1i)−(4−7i)∣=∣−6+8i∣=36+64​=10.

Flashcard 12: Find the midpoint of z1=1+2iz_1=1+2iz1​=1+2i and z2=9+10iz_2=9+10iz2​=9+10i.

Answer: 5+6i5+6i5+6i. (1+2i)+(9+10i)2=10+12i2=5+6i\frac{(1+2i)+(9+10i)}{2} = \frac{10+12i}{2} = 5+6i2(1+2i)+(9+10i)​=210+12i​=5+6i.

Flashcard 13: What is the distance between z1z_1z1​ and z2z_2z2​ if z1−z2=3−4iz_1-z_2=3-4iz1​−z2​=3−4i?

Answer: 555. ∣3−4i∣=9+16=5|3-4i| = \sqrt{9+16} = 5∣3−4i∣=9+16​=5.

Flashcard 14: Identify the distance between z1z_1z1​ and z2z_2z2​ if z1−z2=0+7iz_1-z_2=0+7iz1​−z2​=0+7i.

Answer: 777. ∣0+7i∣=∣7i∣=7|0+7i| = |7i| = 7∣0+7i∣=∣7i∣=7.

Flashcard 15: Identify the distance between z1z_1z1​ and z2z_2z2​ if z1−z2=−9+0iz_1-z_2=-9+0iz1​−z2​=−9+0i.

Answer: 999. ∣−9+0i∣=∣−9∣=9|-9+0i| = |-9| = 9∣−9+0i∣=∣−9∣=9.

Flashcard 16: Find the midpoint of z1=2−4iz_1=2-4iz1​=2−4i and z2=−6+0iz_2=-6+0iz2​=−6+0i.

Answer: −2−2i-2-2i−2−2i. (2−4i)+(−6+0i)2=−4−4i2=−2−2i\frac{(2-4i)+(-6+0i)}{2} = \frac{-4-4i}{2} = -2-2i2(2−4i)+(−6+0i)​=2−4−4i​=−2−2i.

Flashcard 17: Find the distance between z1=2−4iz_1=2-4iz1​=2−4i and z2=−6+0iz_2=-6+0iz2​=−6+0i.

Answer: 80\sqrt{80}80​. ∣(2−4i)−(−6+0i)∣=∣8−4i∣=64+16=80|(2-4i)-(-6+0i)| = |8-4i| = \sqrt{64+16} = \sqrt{80}∣(2−4i)−(−6+0i)∣=∣8−4i∣=64+16​=80​.

Flashcard 18: Find the distance between z1=3+0iz_1=3+0iz1​=3+0i and z2=0+4iz_2=0+4iz2​=0+4i.

Answer: 555. ∣(3+0i)−(0+4i)∣=∣3−4i∣=9+16=5|(3+0i)-(0+4i)| = |3-4i| = \sqrt{9+16} = 5∣(3+0i)−(0+4i)∣=∣3−4i∣=9+16​=5.

Flashcard 19: Find the midpoint of z1=3+0iz_1=3+0iz1​=3+0i and z2=0+4iz_2=0+4iz2​=0+4i.

Answer: 32+2i\frac{3}{2}+2i23​+2i. (3+0i)+(0+4i)2=3+4i2=32+2i\frac{(3+0i)+(0+4i)}{2} = \frac{3+4i}{2} = \frac{3}{2}+2i2(3+0i)+(0+4i)​=23+4i​=23​+2i.

Flashcard 20: Identify the real part of the midpoint m=z1+z22m=\frac{z_1+z_2}{2}m=2z1​+z2​​ for z1=a+biz_1=a+biz1​=a+bi and z2=c+diz_2=c+diz2​=c+di.

Answer: a+c2\frac{a+c}{2}2a+c​. Average the real parts: a+c2\frac{a+c}{2}2a+c​.

Flashcard 21: Identify the imaginary part of the midpoint m=z1+z22m=\frac{z_1+z_2}{2}m=2z1​+z2​​ for z1=a+biz_1=a+biz1​=a+bi and z2=c+diz_2=c+diz2​=c+di.

Answer: b+d2\frac{b+d}{2}2b+d​. Average the imaginary parts: b+d2\frac{b+d}{2}2b+d​.

Flashcard 22: What is the midpoint of z1=az_1=az1​=a and z2=bz_2=bz2​=b when both endpoints are real numbers?

Answer: a+b2\frac{a+b}{2}2a+b​. For real numbers, the midpoint formula reduces to the average.

Flashcard 23: What is the distance between z1=az_1=az1​=a and z2=bz_2=bz2​=b when both endpoints are real numbers?

Answer: ∣a−b∣|a-b|∣a−b∣. For real numbers, distance is the absolute value of the difference.

Flashcard 24: Find the distance between z1=5z_1=5z1​=5 and z2=−1+4iz_2=-1+4iz2​=−1+4i.

Answer: 52\sqrt{52}52​. ∣5−(−1+4i)∣=∣6−4i∣=36+16=52|5-(-1+4i)| = |6-4i| = \sqrt{36+16} = \sqrt{52}∣5−(−1+4i)∣=∣6−4i∣=36+16​=52​.

Flashcard 25: Find the midpoint of z1=−3iz_1=-3iz1​=−3i and z2=9iz_2=9iz2​=9i.

Answer: 3i3i3i. −3i+9i2=6i2=3i\frac{-3i+9i}{2} = \frac{6i}{2} = 3i2−3i+9i​=26i​=3i.

Flashcard 26: Find the midpoint of z1=5z_1=5z1​=5 and z2=−1+4iz_2=-1+4iz2​=−1+4i.

Answer: 2+2i2+2i2+2i. 5+(−1+4i)2=4+4i2=2+2i\frac{5+(-1+4i)}{2} = \frac{4+4i}{2} = 2+2i25+(−1+4i)​=24+4i​=2+2i.

Flashcard 27: State the formula for the distance between complex numbers z1z_1z1​ and z2z_2z2​ in the complex plane.

Answer: d=∣z1−z2∣d=|z_1-z_2|d=∣z1​−z2​∣. Distance is the modulus of the difference between two complex numbers.

Flashcard 28: State the formula for the midpoint of endpoints z1z_1z1​ and z2z_2z2​ in the complex plane.

Answer: m=z1+z22m=\frac{z_1+z_2}{2}m=2z1​+z2​​. Average the two endpoints by adding and dividing by 2.

Flashcard 29: What is the modulus of z=a+biz=a+biz=a+bi written in terms of aaa and bbb?

Answer: ∣z∣=a2+b2|z|=\sqrt{a^2+b^2}∣z∣=a2+b2​. Modulus uses the Pythagorean theorem with real and imaginary parts.

Flashcard 30: What is the distance from the origin to zzz in the complex plane, written using modulus?

Answer: ∣z∣|z|∣z∣. The modulus gives the distance from any point to the origin.