# Eureka Math Geometry Module 4 Lesson 15 Answer Key

## 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?
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?
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
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

c. p(0, 0) and the line y = x – 6

### 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.
The slope is – 1.

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

Question 3.
Confirm the results obtained in Problem 2 using another method.
$$\overline{P R}$$ 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.

∆ 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.