## Engage NY Eureka Math Geometry Module 4 Lesson 7 Answer Key

### Eureka Math Geometry Module 4 Lesson 7 Example Answer Key

Given A(5, -7) and B(8, 2):

a. Find an equation for the line through A and perpendicular to \(\overline{A B}\).

Answer:

A'(0, 0), B'(8 – 5, 2 – (-7)), and P'(x – 5, y – (-7))

(8 – 5) (x – 5) + (2 + 7) (y + 7) = 0

3(x – 5) = -9(y + 7)

3x – 15 = – 9y – 63

y = –\(\frac{1}{3}\)x –\(\frac{16}{3}\)

b. Find an equation for the line through B and perpendicular to \(\overline{A B}\).

Answer:

B’(0, 0), A'(5 – 8,- 7 – 2), and P'(x – 8, y – 2)

(5 – 8) (x – 8) + (- 7 – 2) (y – 2) = 0

– 3(x – 8) = 9(y – 2)

– 3x + 24 = 9y – 18

y = –\(\frac{1}{3}\)x –\(\frac{14}{3}\)

### Eureka Math Geometry Module 4 Lesson 7 Opening Exercise Answer Key

The equations given are in standard form. Put each equation In slope-intercept form. State the slope and the y-intercept.

Exercise 1.

6x + 3y = 12

Answer:

y = – 2x + 4

slope = – 2

y-intercept = 4

Exercise 2.

5x + 7y = 14

Answer:

y = –\(\frac{5}{7}\)x + 2

slope = –\(\frac{5}{7}\)

y-intercept = 2

3.

2x – 5y = – 7

Answer:

y = \(\frac{2}{5}\)x + \(\frac{7}{5}\)

slope = \(\frac{2}{5}\)

y-intercept = \(\frac{7}{5}\)

### Eureka Math Geometry Module 4 Lesson 7 Exercise Answer Key

Exercise 1.

Given U(- 4, – 1) and V(7, 1):

a. Write an equation for the line through U and perpendicular to \(\overline{U V}\).

Answer:

U'(0, 0), V'(7 – (-4), 1 – (-1)), P'(x – (-4), y – (-1))

(7 + 4) (x + 4) + (1 + 1) (y + 1) = 0

11x + 44 = – 2y – 2

y = –\(\frac{11}{2}\)x – 23

b. Write an equation for the line through V and perpendicular to \(\overline{U V}\).

Answer:

V'(0, 0), U'(-4 – 7,-1 – 1), P'(x – 7, y – 1)

(-7 – 4) (x – 7) + (-1 – 1) (y – 1) = 0

-11x + 77 = 2y – 2

y = –\(\frac{11}{2}\)x + \(\frac{79}{2}\)

Exercise 2.

Given S(5, – 4) and T(-8, 12):

a. Write an equation for the line through S and perpendicular to \(\overline{S T}\).

Answer:

S'(0, 0), T'(-8 -5, 12 – (-4)), P'(x – 5, y – (-4))

(-8 – 5) (x – 5) + (12 – (-4)) (y – (-4)) = 0

– 13x + 65 = – 16y – 64

y = \(\frac{13}{16}\)x – \(\frac{129}{16}\)

b. Write an equation for the line through T and perpendicular to \(\overline{S T}\).

Answer:

T'(0, 0), S'(5-(-8), – 4 – 12), P'(x – (-8), y – 12)

(5 – (-8)) (x – (-8)) + (-4 – 12) (y – 12) = 0

13x + 104 = 16y – 192

y = \(\frac{13}{16}\)x – \(\frac{37}{2}\)

### Eureka Math Geometry Module 4 Lesson 7 Problem Set Answer Key

Question 1.

Given points C(-4, 3) and D(3, 3):

a. Write the equation of the line through C and perpendicular to \(\overline{C D}\).

Answer:

x = – 4

b. Write the equation of the line through D and perpendicular to \(\overline{C D}\)

Answer:

x = 3

Question 2.

Given points N(7, 6) and M(7, -2):

a. Write the equation of the line through M and perpendicular to \(\overline{M N}\).

Answer:

y = – 2

b. Write the equation of the line through N and perpendicular to\(\overline{M N}\).

Answer:

y = 6

Question 3.

The equation of a line is given by the equation 8(x – 4) + 3(y + 2) = 0.

a. What are the coordinates of the image of the endpoint of the normal segment that does not lie on the line? Explain your answer.

Answer:

(8, 3) because A(x – 4) + B(y + 2) = 0 is the original formula, and (A, B) are the coordinates of the image of the endpoint of the normal segment not on the line.

b. What translation occurred to move the point of perpendicularity to the origin?

Answer:

(- 4, 2), or left 4 units up 2 units

c. What were the coordinates of the original point of perpendicularity? Explain your answer.

Answer:

(4, – 2) because the translation of (- 4, 2) was required to move the point of perpendicularity to the origin.

d. What were the endpoints of the original normal segment?

Answer:

8 = c – 4 and 3 = d – (-2)

c = 12 d = 1

The endpoints of a normal segment to the given line are A(4, – 2) and B(12, 1).

Question 4.

A coach is laying out lanes for a race. The lanes are perpendicular to a segment of the track such that one endpoint of the segment is (2, 50), and the other Is (20, 65). What are the equations of the lines through the endpoints?

Answer:

y = –\(\frac{6}{5}\)x + \(\frac{786}{15}\);y = –\(\frac{6}{5}\)x + 89

### Eureka Math Geometry Module 4 Lesson 7 Exit Ticket Answer Key

Given A(-5, -3), B(-1, 6), and C(x, y):

a. What are the coordinates of the translated points if B moves to the origin?

Answer:

A'(-4, -9), B'(0, 0), C'(x + 1, y – 6)

b. Write the condition for perpendicularity equation.

Answer:

-4(x + 1) – 9 (y – 6) = 0

c. Write the equation for the normal line in slope-intercept form.

Answer:

y = –\(\frac{4}{9}\)x + \(\frac{50}{9}\)

### Eureka Math Geometry Module 4 Lesson 7 Closing Answer Key

Describe the characteristics of a normal segment.

Answer:

A line segment with one endpoint on a line and perpendicular to the line is called a normal segment to the line.

Every equation of a line through a given point (a, b) has the form A(x – a) + B(y – b) = 0. Explain how the values of A and B are obtained.

Answer:

A is the value of the abscissa of the image of the endpoint of the normal segment that does not lie on the line after the figure has been translated so that the image of the endpoint that does lie on the line, the point of perpendicularity, is at the origin.

B is the value of the ordinate of the image of the endpoint of the normal segment that does not lie on the line after the figure has been translated so that the image of the endpoint that does lie on the line, the point of perpendicularity, is at the origin.