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 = $$\sqrt{1300}$$ = 10âˆš13

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 âˆ† ABC with vertices A(3, -1), B(2, 2), and C(5, 1):
a. Find the slope of the angle bisector of âˆ ABC.
The slope is – 1.

b. Prove that the bisector of âˆ ABC is the perpendicular bisector of $$\overline{A C}$$.
Let $$\overline{B D}$$ be the bisector of âˆ ABC, where D is the point of Intersection with $$\overline{A C}$$.
AB = CB = âˆš10; therefore, âˆ† ABC is isosceles, and mâˆ A = mâˆ C (base angles of isosceles have equal measures).
mâˆ ABD = mâˆ CBD by definition of angle bisector.
âˆ† ABD â‰… âˆ† 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, $$\overline{B D}$$ âŠ¥ $$\overline{A C}$$.
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 âˆ ABC intersects $$\overline{A C}$$.
y = – x + 4

Question 2.
Use the distance formula from todayâ€™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 = $$\sqrt{2^{2}+1^{2}}$$ = âˆš5.

Question 4.
Find the perimeter of quadrilateral DEBF, shown below.

âˆ† AED is a right triangle with hypotenuse $$\overline{A D}$$ with length 8 and leg $$\overline{E D}$$ with length 4âˆš2. This means that âˆ† AED is an isosceles right triangle (45Â° – 45Â° – 90Â°), which means AE = 4âˆš2 We also know that âˆ† AED â‰… âˆ† DEB â‰… âˆ† BFD â‰… âˆ† DFC because they are all right triangles with hypotenuse of length 8 and leg of length 4âˆš2. Therefore, the perimeter of DEBF is 4(4âˆš2) = 16âˆš2.

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 todayâ€™s lesson.