## Engage NY Eureka Math Precalculus Module 1 Lesson 11 Answer Key

### Eureka Math Precalculus Module 1 Lesson 11 Exercise Answer Key

Opening Exercise

a. Plot the complex number z=2+3i on the complex plane. Plot the ordered pair (2,3) on the coordinate plane.

b. In what way are complex numbers points?
When a complex number is plotted on a complex plane, it looks just like the corresponding ordered pair plotted on a coordinate plane. For example, when 2+3i is plotted on the complex plane, it looks exactly the same as when the ordered pair (2,3) is plotted on a coordinate plane. We can interchangeably think of a complex number x+yi in the complex plane as a point (x,y) in the coordinate plane, and vice versa.

c. What point on the coordinate plane corresponds to the complex number -1+8i?
(-1,8)

d. What complex number corresponds to the point located at coordinate (0,-9)?
0-9i or -9i

Exercise 1.
The endpoints of $$\overline{\boldsymbol{AB}}$$ are A(1,8) and B(-5,3). What is the midpoint of $$\overline{\boldsymbol{AB}}$$?
The midpoint of $$\overline{\boldsymbol{AB}}$$ is (-2,$$\frac{11}{2}$$).

→ How do you find the midpoint of $$\overline{\boldsymbol{AB}}$$?
→ You find the average of the x-coordinates and the y-coordinates to find the halfway point.
→ In Geometry, we learned that for two points A(x1,y1) and B(x2, y2), the midpoint of $$\overline{\boldsymbol{AB}}$$ is ($$\frac{x_{1}+x_{2}}{2}$$, $$\frac{y_{1}+y_{2}}{2}$$).
→ Now, view these points as complex numbers: A=x1+y1i and B=x2 + y2i.

Exercise 2.
What is the midpoint of A=1+8i and B=-5+3i?
M=$$\frac{1+-5}{2}$$ + $$\frac{(8+3)}{2}$$i = -2 + $$\frac{11}{2}$$i

b. Using A=x1+y1i and B=x2+y2i, show that, in general, the midpoint of points A and B is $$\frac{\boldsymbol{A}+\boldsymbol{B}}{2}$$, the arithmetic average of the two numbers.
M=$$\frac{x_{1}+x_{2}}{2}$$+$$\frac{y_{1}+y_{2}}{2}$$i
= $$\frac{x_{1}+y_{1} i+x_{2}+y_{2} i}{2}$$
= $$\frac{\boldsymbol{A}+\boldsymbol{B}}{2}$$

Exercise 3.
The endpoints of $$\overline{\boldsymbol{AB}}$$ are A(1,8) and B(-5,3). What is the length of $$\overline{\boldsymbol{AB}}$$?
The length of $$\overline{\boldsymbol{AB}}$$ is √61.

→ How do you find the length of $$\overline{\boldsymbol{AB}}$$?
→ You use the Pythagorean theorem, which can be written as AB=$$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$ for two points A(x1, y1) and B(x2, y2).
→ As we did previously, view these points as complex numbers: A=x1+y1i and B=x2+y2i.

Exercise 4.
a. What is the distance between A=1+8i and B=-5+3i?
d=$$\sqrt{(1-(-5))^{2}+(8-3)^{2}}$$=$$\sqrt{61}$$

b. Show that, in general, the distance between A=x1+y1i and B=x2+y2i is the modulus of A-B.
A-B=(x1 – x2)+(y1 – y2)i
|A-B|=$$\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$$= distance between A and B

Exercise 5.
Suppose z=2+7i and w=-3+i.

a. Find the midpoint m of z and w.
m=-$$\frac{1}{2}$$+4i

b. Verify that |z-m|=|w-m|.

### Eureka Math Precalculus Module 1 Lesson 11 Problem Set Answer Key

Question 1.
Find the midpoint between the two given points in the rectangular coordinate plane.

a. 2+4i and 4+8i
M=$$\frac{2+4}{2}$$+$$\frac{4+8}{2}$$i=3+6i

b. -3+7i and 5-i
M=$$\frac{-3+5}{2}$$+$$\frac{7-1}{2}$$i=1+3i

c. -4+3i and 9-4i
M=$$\frac{-4+9}{2}$$+$$\frac{3-4}{2}$$i=$$\frac{5}{2}$$–$$\frac{1}{2}$$ i

d. 4+i and -12-7i
M=$$\frac{4-12}{2}$$+$$\frac{1-7}{2}$$i=-4-3i

e. -8-3i and 3-4i
M=$$\frac{-8+3}{2}$$+$$\frac{-3-4}{2}$$i=-$$\frac{5}{2}$$ – $$\frac{7}{2}$$i

f. $$\frac{2}{3}$$–$$\frac{5}{2}$$i and -0.2+0.4i

Question 2.
Let A=2+4i, B=14+8i, and suppose that C is the midpoint of A and B and that D is the midpoint of A and C.
a. Find points C and D.

b. Find the distance between A and B.
|A-B|=|2+4i-14-8i|
=|-12-4i|
=$$\sqrt{(-12)^{2}+(-4)^{2}}$$
=$$\sqrt{144+16}$$
=$$\sqrt{160}$$
=$$4 \sqrt{10}$$

c. Find the distance between A and C.
|A-C|=|2+4i-8-6i|
=|-6-2i|
=$$\sqrt{(-6)^{2}+(-2)^{2}}$$
=$$\sqrt{40}$$
=$$2 \sqrt{10}$$

d. Find the distance between C and D.
|C-D|=|8+6i-5-5i|
=|3+i|
=$$\sqrt{(3)^{2}+(1)^{2}}$$
=$$\sqrt{10}$$

e. Find the distance between D and B.
|D-B|=|5+5i-14-8i|
=|-9-3i|
=$$\sqrt{(-9)^{2}+(-3)^{2}}$$
=$$\sqrt{90}$$
=3$$\sqrt{10}$$

f. Find a point one-quarter of the way along the line segment connecting segment A and B, closer to A than
to B.
The point is D=5+5i.

g. Terrence thinks the distance from B to C is the same as the distance from A to B. Is he correct? Explain why or why not.
The distance from B to C is 2$$\sqrt{10}$$, and the distance from A to B is 4$$\sqrt{10}$$. The distances are not the same.

h. Using your answer from part (g), if E is the midpoint of C and B, can you find the distance from E to C? Explain.
The distance from B to C is 2$$\sqrt{10}$$, and the distance from E to C should be half of this value, $$\sqrt{10}$$.

i. Without doing any more work, can you find point E? Explain.
B is 5+5i, which is 3 units to the right of A in the real direction and 1 unit up in the imaginary direction. From C, you should move the same amount to get to E, so E would be 11+7i.

### Eureka Math Precalculus Module 1 Lesson 11 Exit Ticket Answer Key

Question 1.
Kishore said that he can add two points in the coordinate plane like adding complex numbers in the complex plane. For example, for point A(2,3) and point B(5,1), he will get A+B=(7,4). Is he correct? Explain your reasoning.
No. Kishore is not correct because we cannot add two points in the rectangular plane. However, we can add two complex numbers in the complex plane, which has the geometric effect of performing a translation to points in complex numbers.

Question 2.
Consider two complex numbers A=-4+5i and B=4-10i.
a. Find the midpoint of A and B.
M=$$\frac{A+B}{2}$$
=$$\frac{-4+5 i+4-10 i}{2}$$
=$$\frac{-5 i}{2}$$
=$$-\frac{5}{2} i$$ or 0-$$\frac{5}{2}$$i
d=$$\sqrt{(4-(-4))^{2}+(-10-5)^{2}}$$=17