## Engage NY Eureka Math Geometry Module 4 Lesson 15 Answer Key

### Eureka Math Geometry Module 4 Lesson 15 Exercise Answer Key

Exercise 1.

A robot is moving along the line 20x + 30y = 600. A homing beacon sits at the point (35, 40).

a. Where on this line will the robot hear the loudest ping?

Answer:

Students need to determine the equation of the line passing through the point (35, 40) that is perpendicular to the line 20x + 30y = 600. The slope of this line is –\(\frac{2}{3}\).

y – 40 = (x – 35) or y = (x – 35) + 40

There are a variety of methods available to students to use to determine the point where these two lines intersect. Using substitution is one of the more efficient methods.

20x + 30 (\(\frac{3}{2}\)(x – 35) + 40) = 600

20x + 45x – 1575 + 1200 = 600

65x = 975

x = 15

y = \(\frac{3}{2}\)(15 – 35) + 40

y = 10

The robot will be closest to the beacon when it is on the point (15, 10).

b. At this point, how far will the robot be from the beacon?

Answer:

Students need to calculate the distance between the two points B(35, 40) and A(15, 10). Encourage students to think about the right triangle that is created if one moves from A to B in two moves: one horizontal and the other vertical.

AB = \(\sqrt{(35-15)^{2}+(40-10)^{2}}\)

AB = \(\sqrt{(20)^{2}+(30)^{2}}\)

AB = \(\sqrt{1300}\) = 10√13

Exercise 2.

For the following problems, use the formula to calculate the distance between the point P and the line 1.

a. p(0, 0) and the line y = 10

Answer:

p = 0, q = 0, m = 0, and b = 10

d = \(\sqrt{\left(\frac{0+0-10(0)}{1+0^{2}}-0\right)^{2}+\left(0\left(\frac{0+0-10(0)}{1+0^{2}}\right)+10-0\right)^{2}}\)

d = \(\sqrt{0+(10)^{2}}\)

d = 10

b. p(0, 0) and the line y = x + 10

Answer:

c. p(0, 0) and the line y = x – 6

Answer:

### Eureka Math Geometry Module 4 Lesson 15 Problem Set Answer Key

Question 1.

Given ∆ ABC with vertices A(3, -1), B(2, 2), and C(5, 1):

a. Find the slope of the angle bisector of ∠ABC.

Answer:

The slope is – 1.

b. Prove that the bisector of ∠ABC is the perpendicular bisector of \(\overline{A C}\).

Answer:

Let \(\overline{B D}\) be the bisector of ∠ABC, where D is the point of Intersection with \(\overline{A C}\).

AB = CB = √10; therefore, ∆ ABC is isosceles, and m∠A = m∠C (base angles of isosceles have equal measures).

m∠ABD = m∠CBD by definition of angle bisector.

∆ ABD ≅ ∆ CBD byASA.

BD = CD, since corresponding sides of congruent triangles have equal lengths; therefore, \(\overline{B D}\) bisects \(\overline{A C}\).

The slope of \(\overline{B D}\) is -1; therefore, \(\overline{B D}\) ⊥ \(\overline{A C}\).

Therefore, \(\overline{B D}\) is the perpendicular bisector of \(\overline{A C}\).

c. Write the equation of the line containing \(\overline{B D}\), where point D is the point where the bisector of ∠ABC intersects \(\overline{A C}\).

Answer:

y = – x + 4

Question 2.

Use the distance formula from today’s lesson to find the distance between the point P(-2, 1) and the line y = 2x.

Answer:

Question 3.

Confirm the results obtained in Problem 2 using another method.

Answer:

\(\overline{P R}\) is the hypotenuse of the right triangle with vertices P(-2, 1), R(0, 0), and (-2, 0). Using the Pythagorean theorem, we find that PR = \(\sqrt{2^{2}+1^{2}}\) = √5.

Question 4.

Find the perimeter of quadrilateral DEBF, shown below.

Answer:

∆ AED is a right triangle with hypotenuse \(\overline{A D}\) with length 8 and leg \(\overline{E D}\) with length 4√2. This means that ∆ AED is an isosceles right triangle (45° – 45° – 90°), which means AE = 4√2 We also know that ∆ AED ≅ ∆ DEB ≅ ∆ BFD ≅ ∆ DFC because they are all right triangles with hypotenuse of length 8 and leg of length 4√2. Therefore, the perimeter of DEBF is 4(4√2) = 16√2.

### Eureka Math Geometry Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.

Find the distance between the point P(0, 0) and the line y = – x + 4 using the formula from today’s lesson.

Answer:

Question 2.

Verify using another method.

Answer: