Eureka Math Geometry Module 4 Lesson 6 Answer Key

Engage NY Eureka Math Geometry Module 4 Lesson 6 Answer Key

Eureka Math Geometry Module 4 Lesson 6 Example Answer Key

Example
Given points A(2, 2), B(10, 16), C( – 3, 1), and D(4, – 3), are $$\overline{A B}$$ and $$\overline{C D}$$ perpendicular? Are the lines containing the segments perpendicular? Explain.
One possible solution would be to translate $$\overline{A B}$$ so that A’ is on the origin (using the vector ( – 2, – 2), or left 2 units and down 2 units) and to translate $$\overline{C D}$$ so that C’ is on the origin (using the vector (3, – 1), right 3 units and down 1 unit):

A(2, 2) → A'(2 – 2, 2 – 2) → A'(0, 0)
B(10, 16) → B'(10 – 2, 16 – 2) → B'(8, 14)
C( – 3, 1) → C'( – 3 – ( – 3), 1 – 1) → C'(0, 0)
D(4, – 3) → D'(4 – ( – 3), – 3 – 1) → D'(7, – 4)
$$\overline{A B}$$ will be perpendicular to $$\overline{C D}$$ if $$\overline{A^{\prime} B^{\prime}}$$ is perpendicular to $$\overline{C^{\prime} D^{\prime}}{\prime}$$.
8(7) + 14( – 4) = 0; therefore, $$\overline{A^{\prime} B^{\prime}}$$ is perpendicular to $$\overline{C^{\prime} D^{\prime}}$$, which means $$\overline{A B}$$ ⊥ $$\overline{C D}$$.

Eureka Math Geometry Module 4 Lesson 6 Exercise Answer Key

Opening Exercise
Carlos thinks that the segment having endpoints A(0, 0) and B(6, 0) is perpendicular to the segment with endpoints A(0, 0) and C( – 2, 0). Do you agree? Why or why not?
No, the two segments are not perpendicular. If they were perpendicular, then 6( – 2) + 0(0) = 0 would be true.

Working with a partner, given A(0, 0) and B(3, – 2), find the coordinates of a point C so that $$\overline{A C}$$ ⊥ $$\overline{A B}$$.
Let the other endpoint be C(c, d). If $$\overline{A C}$$ ⊥ $$\overline{A B}$$, then 3c + – 2d = 0. This means that $$\overline{A C}$$ will be perpendicular to $$\overline{A B}$$ as long as we choose values for c and d that satisfy the equation d = $$\frac{1}{2}$$ c.
Answers may vary but may include (2, 3), (4, 6), and (6, 9) or any other coordinates that meet the requirement stated above.

Exercises
Exercise 1.
Given A(a1, a2), B(b1, b2), C(c1, c2), and D(d1, d2), find a general formula in terms of a1, a2, b1, b2, c1, c2, d1, and d2 that will let us determine whether $$\overline{A B}$$ and $$\overline{C D}$$ are perpendicular.
After translating the segments so that the image of points A and C lie on the origin, we get A’ (0, 0), B’ (b1 – a1, b2 – a2 ), C’ (0, 0), and D’ (d1 – c1, d2 – c2 ).
$$\overline{A^{\prime} B^{\prime}}$$ ⊥ $$\overline{A^{\prime} B^{\prime}}$$⟺ (b1 – a1 )(d1 – c1 ) + (b2 – a2 )(d2 – c2 ) = 0

Exercise 2.
Recall the Opening Exercise of Lesson 4 in which a robot is traveling along a linear path given by the equation
y = 3x – 600. The robot hears a ping from a homing beacon when it reaches the point F(400, 600) and turns to travel along a linear path given by the equation y – 600 = – $$\frac{1}{3}$$(x – 400). If the homing beacon lies on the x – axis, what is its exact location? (Use your own graph paper to visualize the scenario.)

a. If point E is the y – intercept of the original equation, what are the coordinates of point E?
E(0, – 600)

b. What are the endpoints of the original segment of motion?
E(0, – 600) and F(400, 600)

c. If the beacon lies on the x – axis, what is the y – value of this point, G?
0

d. Translate point F to the origin. What are the coordinates of E’, F’, and G’?
E'( – 400, – 1200), F’ (0, 0), and G'(g – 400, – 600)

e. Use the formula derived in this lesson to determine the coordinates of point G.
We know that $$\overline{E F}$$ ⊥ $$\overline{F G}$$, so – 400(g – 400) + ( – 1200)( – 600) = 0 ⟺ g = 2200.
Given G(g, 0) and we found g = 2000, then the coordinates of point G are (2200, 0).

Exercise 3.
A triangle in the coordinate plane has vertices A(0, 10), B( – 8, 8), and C( – 3, 5). Is it a right triangle? If so, at which vertex is the right angle? (Hint: Plot the points, and draw the triangle on a coordinate plane to help you determine which vertex is the best candidate for the right angle.)
The most likely candidate appears to be vertex C. By translating the figure so that C is mapped to the origin, we can use the formula

( – 8 + 3)(0 + 3) + (8 – 5)(10 – 5) = 0
– 5(3) + 3(5) = 0
0 = 0
Yes, $$\overline{B C}$$ and $$\overline{A C}$$ are perpendicular. The right angle is ∠C.

Exercise 4.
A( – 7, 1), B( – 1, 3), C(5, – 5), and D( – 5, – 5) are vertices of a quadrilateral. If $$\overline{A C}$$ bisects $$\overline{B D}$$, but $$\overline{B D}$$does not bisect $$\overline{A C}$$, determine whether ABCD is a kite.
We can show ABCD is a kite if $$\overline{A C}$$ ⊥ $$\overline{B D}$$
Translating $$\overline{A C}$$ so that the image of point A lies on the origin and translating $$\overline{B D}$$ so that the image of point B lies at the origin lead to the equation:
(5 + 7)( – 5 + 1) + ( – 5 – 1)( – 5 – 3) = 0
(12)( – 4) + ( – 6)( – 8) = 0
– 48 + 48 = 0
This is a true statement; therefore, $$\overline{A C}$$ ⊥ $$\overline{B D}$$, and ABCD is a kite.

Eureka Math Geometry Module 4 Lesson 6 Problem Set Answer Key

Question 1.
Are the segments through the origin and the points listed perpendicular? Explain.
a. A(9, 10), B(10, 9)
No 9∙10 + 10∙9 ≠ 0

b. C(9, 6), D(4, – 6)
Yes 9∙4 + 6∙( – 6) = 0

Question 2.
Given M(5, 2), N(1, – 4), and L listed below, are $$\overline{L M}$$ and $$\overline{M N}$$ perpendicular? Translate M to the origin, write the coordinates of the images of the points, and then explain without using slope.
a. L( – 1, 6)
M'(0, 0), N'( – 4, – 6), L'( – 6, 4)
Yes ( – 4)( – 6) + ( – 6)(4) = 0

b. L(11, – 2)
M'(0, 0), N'( – 4, – 6), L'(6, – 4)
Yes ( – 4)(6) + ( – 6)( – 4) = 0

c. L(9, 8)
M'(0, 0), N'( – 4, – 6), L'(4, 6)
No. ( – 4)(4) + ( – 6)(6) ≠ 0

Question 3.
Is triangle PQR, where P( – 7, 3), Q( – 4, 7), and R(1, – 3), a right triangle? If so, which angle is the right angle? Justify your answer.
Yes. If the points are translated to P'(0, 0), Q'(3, 4), and R'(8, – 6), 3(8) + 4( – 6) = 0, meaning the segments are perpendicular. The right angle is ∠P.

Question 4.
A quadrilateral has vertices (2 + $$\sqrt{2}$$, – 1), (8 + $$\sqrt{2}$$, 3), (6 + $$\sqrt{2}$$, 6), and ($$\sqrt{2}$$, 2). Prove that the quadrilateral is a rectangle.
Answers will vary, but it is a rectangle because it has 4 right angles.

Question 5.
Given points G( – 4, 1), H(3, 2), and I( – 2, – 3), find the x – coordinate of point J with y – coordinate 4 so that the $$\overleftrightarrow{G H}$$ and $$\overleftrightarrow{I J}$$ are perpendicular.
– 3

Question 6.
A robot begins at position ( – 80, 45) and moves on a path to (100, – 60). It turns 90° counterclockwise.
a. What point with y – coordinate 120 is on this path?
(205, 120)

b. Write an equation of the line after the turn.
y + 60 = $$\frac{12}{7}$$ (x – 100)

c. If it stops to charge on the x – axis, what is the location of the charger?
(135, 0)

Question 7.
Determine the missing vertex of a right triangle with vertices (6, 2) and (5, 5) if the third vertex is on the y – axis. Verify your answer by graphing.
(0, $$\frac{10}{3}$$)

Question 8.
Determine the missing vertex for a rectangle with vertices (3, – 2), (5, 2), and ( – 1, 5), and verify by graphing. Then, answer the questions that follow.
( – 3, 1)

a. What is the length of the diagonal?
Approximately 8.06 units

b. What is a point on both diagonals in the interior of the figure?
(1, $$\frac{3}{2}$$)

Question 9.
Leg $$\overline{A B}$$ of right triangle ABC has endpoints A(1, 3) and B(6, – 1). Point C(x, y) is located in Quadrant IV.
a. Use the perpendicularity criterion to determine at which vertex the right angle is located. Explain your reasoning.
Assume that the right angle is at A(1, 3). Then a translation 1 unit left and 3 units down maps point A to the origin. A(1, 3) → A’ (1 – 1, 3 – 3) → A’ (0, 0)
B(6, – 1) → B'(6 – 1, – 1 – 3) → B'(5, – 4)
C(x, y) → C'(x – 1, y – 3)
By the criterion for perpendicularity,
5(x – 1) + – 4(y – 3) = 0
5x – 5 – 4y + 12 = 0
5x – 4y + 7 = 0
4y = 5x + 7
y = $$\frac{5}{4}$$ x + $$\frac{7}{4}$$
The solutions to this equation form a line that had a positive slope and a positive y – intercept so no point on this line lies in Quadrant IV. Therefore, the right angle cannot be at A(1, 3) and so must be at B(6, – 1).

b. Determine the range of values that x is limited to and why?
If the right angle is at B(6, – 1), then a translation 6 units left and 1 unit up maps point B to the origin.
B(6, – 1) → B'(6 – 6, – 1 – ( – 1)) → B'(0, 0)
A(1, 3) → A'(1 – 6, 3 – ( – 1)) → A'( – 5, 4)
C(x, y) → C'(x – 6, y – ( – 1)) → C'(x – 6, y + 1)
By the criterion for perpendicularity,
– 5(x – 6) + 4(y + 1) = 0
– 5x + 30 + 4y + 4 = 0
– 5x + 4y + 34 = 0
4y = 5x – 34
y = $$\frac{5}{4}$$ x – $$\frac{17}{2}$$
The point must lie on the graph of y = $$\frac{5}{4}$$ x – $$\frac{17}{2}$$ and in Quadrant IV. The y – intercept of the graph is (0, – $$\frac{17}{2}$$), however since this point is not in the fourth quadrant, x > 0. Substituting 0 for y in the equation, it is determined that the x – intercept of the graph is 6.8. This point does not lie in Quadrant IV so x < 6.8. The remaining vertex cannot coincide with another vertex, so x ≠ 6. Therefore, the value of the x – coordinate of the remaining vertex is limited to 0 < x < 6.8 and x≠6.

c. Find the coordinates of point C if they are both integers.
The value of x must be 1, 2, 3, 4, or 5. Substituting each into the equation y = $$\frac{5}{4}$$ x – $$\frac{17}{2}$$, the only value of x that yields an integer value for y is x = 2.
y = $$\frac{5}{4}$$ (2) – $$\frac{17}{2}$$
y = $$\frac{5}{2}$$ – $$\frac{17}{2}$$
y = – 6
The coordinates of vertex C are (2, – 6).

Eureka Math Geometry Module 4 Lesson 6 Exit Ticket Answer Key

Question 1.
Given points S(2, 4), T(7, 6), U( – 3, – 4), and V( – 1, – 9):

a. Translate $$\overline{S T}$$ and $$\overline{U V}$$ so that the image of each segment has an endpoint at the origin.
Answers can vary slightly. Students have to choose one of the first two and one of the second two.
If we translate $$\overline{S T}$$ so that the image of S is at the origin, we get S'(0, 0), T'(5, 2).
If we translate $$\overline{S T}$$ so that the image of T is at the origin, we get S'( – 5, – 2), T'(0, 0).
If we translate $$\overline{U V}$$ so that the image of U is at the origin, we get U'(0, 0), V'(2, – 5).
If we translate $$\overline{U V}$$ so that the image of V is at the origin, we get V'(0, 0), U'( – 2, 5).

b. Are the segments perpendicular? Explain.
Yes. By choosing any two of the translated $$\overline{S^{\prime} T^{\prime}}$$ and $$\overline{U^{\prime} V^{\prime}}$$, we determine whether the equation yields a true statement: (b1 – a1)(d1 – c1 ) + (b2 – a2)(d2 – c2) = 0.
– 5(2) + ( – 2)( – 5) = 0 is a true statement; therefore, $$\overline{S^{\prime} T^{\prime}}$$ ⊥ $$\overline{U^{\prime} V^{\prime}}$$ and $$\overline{S T}$$ ⊥ $$\overline{U V}$$.
c. Are the lines $$\overleftrightarrow{S T}$$ and $$\overleftrightarrow{U V}$$ perpendicular? Explain.