## Engage NY Eureka Math 8th Grade Module 4 Lesson 21 Answer Key

### Eureka Math Grade 8 Module 4 Lesson 21 Example Answer Key

Example 1.

Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:

We can pick two points to determine the slope, but the precise location of the y – intercept point cannot be determined from the graph.

Calculate the slope of the line.

Using points ( – 2, 2) and (5, 4), the slope of the line is

m = \(\frac{2 – 4}{ – 2 – 5}\)

= \(\frac{ – 2}{ – 7}\)

= \(\frac{2}{7}\)

→ Now we need to determine the y – intercept point of the line. We know that it is a point with coordinates (0, b), and we know that the line goes through points ( – 2, 2) and (5, 4) and has slope m = \(\frac{2}{7}\). Using this information, we can determine the coordinates of the y – intercept point and the value of b that we need in order to write the equation of the line.

→ Recall what it means for a point to be on a line; the point is a solution to the equation. In the equation y = mx + b, (x, y) is a solution, and m is the slope. Can we find the value of b? Explain.

Yes. We can substitute one of the points and the slope into the equation and solve for b.

→ Do you think it matters which point we choose to substitute into the equation? That is, will we get a different equation if we use the point ( – 2, 2) compared to (5, 4)?

No, because there can be only one line with a given slope that goes through a point.

→ Verify this claim by using m = \(\frac{2}{7}\) and ( – 2, 2) to find the equation of the line and then by using m = \(\frac{2}{7}\) and (5, 4) to see if the result is the same equation.

Sample student work:

2 = \(\frac{2}{7}\) ( – 2) + b

2 = – \(\frac{4}{7}\) + b

2 + \(\frac{4}{7}\) = – \(\frac{4}{7}\) + \(\frac{4}{7}\) + b

\(\frac{18}{7}\) = b

4 = \(\frac{2}{7}\) (5) + b

4 = \(\frac{10}{7}\) + b

4 – \(\frac{10}{7}\) = \(\frac{10}{7}\) – \(\frac{10}{7}\) + b

\(\frac{18}{7}\) = b

The y – intercept point is at (0, \(\frac{18}{7}\)), and the equation of the line is y = \(\frac{2}{7}\) x + \(\frac{18}{7}\).

→ The equation of the line is

y = \(\frac{2}{7}\) x + \(\frac{18}{7}\).

→ Write it in standard form.

Sample student work:

(y = \(\frac{2}{7}\) x + \(\frac{18}{7}\))7

7y = 2x + 18

– 2x + 7y = 2x – 2x + 18

– 2x + 7y = 18

– 1( – 2x + 7y = 18)

2x – 7y = – 18

Example 2.

Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:

Determine the slope of the line.

Using points ( – 1, 4) and (4, 1), the slope of the line is

m = \(\frac{4 – 1}{ – 1 – 4}\)

= \(\frac{3}{ – 5}\)

= – \(\frac{3}{5}\).

Determine the y – intercept point of the line.

Sample student work:

4 = ( – \(\frac{3}{5}\))( – 1) + b

4 = \(\frac{3}{5}\) + b

4 – \(\frac{3}{5}\) = \(\frac{3}{5}\) – \(\frac{3}{5}\) + b

\(\frac{17}{5}\) = b

The y – intercept point is at (0, \(\frac{17}{5}\)).

→ Now that we know the slope, m = – \(\frac{3}{5}\), and the y – intercept point, (0, \(\frac{17}{5}\)), write the equation of the line l in slope – intercept form.

y = – \(\frac{3}{5}\) x + \(\frac{17}{5}\)

→ Transform the equation so that it is written in standard form.

Sample student work:

y = – \(\frac{3}{5}\) x + \(\frac{17}{5}\)

(y = – \(\frac{3}{5}\) x + \(\frac{17}{5}\))5

5y = – 3x + 17

3x + 5y = – 3x + 3x + 17

3x + 5y = 17

Example 3.

Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:

→ Using points (12, 2) and (13, 7), the slope of the line is

m = \(\frac{2 – 7}{12 – 13}\)

= \(\frac{ – 5}{ – 1}\)

= 5.

→ Now, determine the y – intercept point of the line, and write the equation of the line in slope – intercept form.

Sample student work:

2 = 5(12) + b

2 = 60 + b

b = – 58

The y – intercept point is at (0, – 58), and the equation of the line is y = 5x – 58.

Now that we know the slope, m = 5, and the y – intercept point, (0, – 58), write the equation of the line l in standard form.

Sample student work:

y = 5x – 58

– 5x + y = 5x – 5x – 58

– 5x + y = – 58

– 1( – 5x + y = – 58)

5x – y = 58

Example 4.

Let a line l be given in the coordinate plane. What linear equation is the graph of line l?

Answer:

Using points (3, 1) and ( – 3, – 1), the slope of the line is

m = \(\frac{ – 1 – 1}{ – 3 – 3}\)

= \(\frac{ – 2}{ – 6}\)

= \(\frac{1}{3}\)

The y – intercept point is at (0, 0), and the equation of the line is y = \(\frac{1}{3}\) x.

### Eureka Math Grade 8 Module 4 Lesson 21 Exercise Answer Key

Exercises

Exercise 1.

Write the equation for the line l shown in the figure.

Answer:

Using the points ( – 1, – 3) and (2, – 2), the slope of the line is

m = \(\frac{ – 3 – ( – 2)}{ – 1 – 2}\)

= \(\frac{ – 1}{ – 3}\)

= \(\frac{1}{3}\) .

– 2 = \(\frac{1}{3}\) (2) + b

– 2 = \(\frac{2}{3}\) + b

– 2 – \(\frac{2}{3}\) = \(\frac{2}{3}\) – \(\frac{2}{3}\) + b

– \(\frac{8}{3}\) = b

The equation of the line is y = \(\frac{1}{3}\) x – \(\frac{8}{3}\).

Exercise 2.

Write the equation for the line l shown in the figure.

Answer:

Using the points ( – 3, 7) and (2, 8), the slope of the line is

m = \(\frac{7 – 8}{ – 3 – 2}\)

= \(\frac{ – 1}{ – 5}\)

= \(\frac{1}{5}\).

8 = \(\frac{1}{5}\) (2) + b

8 = \(\frac{2}{5}\) + b

8 – \(\frac{2}{5}\) = \(\frac{2}{5}\) – \(\frac{2}{5}\) + b

\(\frac{38}{5}\) = b

The equation of the line is y = \(\frac{1}{5}\) x + \(\frac{38}{5}\).

Exercise 3.

Determine the equation of the line that goes through points ( – 4, 5) and (2, 3).

Answer:

The slope of the line is

m = \(\frac{5 – 3}{ – 4 – 2}\)

= \(\frac{2}{ – 6}\)

= – \(\frac{1}{3}\)

The y – intercept point of the line is

3 = – \(\frac{1}{3}\) (2) + b

3 = – \(\frac{2}{3}\) + b

\(\frac{11}{3}\) = b.

The equation of the line is y = – \(\frac{1}{3}\) x + \(\frac{11}{3}\).

Exercise 4.

Write the equation for the line l shown in the figure.

Answer:

Using the points ( – 7, 2) and ( – 6, – 2), the slope of the line is

m = \(\frac{2 – ( – 2)}{ – 7 – ( – 6)}\)

= \(\frac{4}{ – 1}\)

= – 4.

– 2 = – 4( – 6) + b

– 2 = 24 + b

– 26 = b

The equation of the line is y = – 4x – 26.

Exercise 5.

A line goes through the point (8, 3) and has slope m = 4. Write the equation that represents the line.

Answer:

3 = 4(8) + b

3 = 32 + b

– 29 = b

The equation of the line is y = 4x – 29.

### Eureka Math Grade 8 Module 4 Lesson 21 Problem Set Answer Key

Question 1.

Write the equation for the line l shown in the figure.

Answer:

Using the points ( – 3, 2) and (2, – 2), the slope of the line is

m = \(\frac{2 – ( – 2)}{ – 3 – 2}\)

= \(\frac{4}{ – 5}\)

= – \(\frac{4}{5}\)

2 = ( – \(\frac{4}{5}\))( – 3) + b

2 = \(\frac{12}{5}\) + b

2 – \(\frac{12}{5}\) = \(\frac{12}{5}\) – \(\frac{12}{5}\) + b

– \(\frac{2}{5}\) = b

The equation of the line is y = – \(\frac{4}{5}\) x – \(\frac{2}{5}\).

Question 2.

Write the equation for the line l shown in the figure.

Answer:

Using the points ( – 6, 2) and ( – 5, 5), the slope of the line is

m = \(\frac{2 – 5}{ – 6 – ( – 5)}\)

= \(\frac{ – 3}{ – 1}\)

= 3.

5 = 3( – 5) + b

5 = – 15 + b

20 = b

The equation of the line is y = 3x + 20.

Question 3.

Write the equation for the line l shown in the figure.

Answer:

Using the points ( – 3, 1) and (2, 2), the slope of the line is

m = \(\frac{1 – 2}{ – 3 – 2}\)

= \(\frac{ – 1}{ – 5}\)

= \(\frac{1}{5}\)

2 = \(\frac{1}{5}\) (2) + b

2 = \(\frac{2}{5}\) + b

2 – \(\frac{2}{5}\) = \(\frac{2}{5}\) – \(\frac{2}{5}\) + b

\(\frac{8}{5}\) = b

The equation of the line is y = \(\frac{1}{5}\) x + \(\frac{8}{5}\).

Question 4.

Triangle ABC is made up of line segments formed from the intersection of lines L_{AB}, L_{BC}, and L_{AC}. Write the equations that represent the lines that make up the triangle.

Answer:

A( – 3, – 3), B(3, 2), C(5, – 2)

The slope of L_{AB}:

m = \(\frac{ – 3 – 2}{ – 3 – 3}\)

= \(\frac{ – 5}{ – 6}\)

= \(\frac{5}{6}\)

2 = \(\frac{5}{6}\) (3) + b

2 = \(\frac{5}{2}\) + b

2 – \(\frac{5}{2}\) = \(\frac{5}{2}\) – \(\frac{5}{2}\) + b

– \(\frac{1}{2}\) = b

The equation of L_{AB} is y = \(\frac{5}{6}\) x – \(\frac{1}{2}\).

The slope of L_{BC}:

m = \( = \frac{2 – ( – 2)}{3 – 5}\)

= \(\frac{4}{ – 2}\)

= – 2

2 = – 2(3) + b

2 = – 6 + b

8 = b

The equation of L_{BC} is y = – 2x + 8.

The slope of L_{AC}:

m = \(\frac{ – 3 – ( – 2)}{ – 3 – 5}\)

= \(\frac{ – 1}{ – 8}\)

= \(\frac{1}{8}\)

– 2 = \(\frac{1}{8}\) (5) + b

– 2 = \(\frac{5}{8}\) + b

– 2 – \(\frac{5}{8}\) = \(\frac{5}{8}\) – \(\frac{5}{8}\) + b

– 2\(\frac{1}{8}\) = b

The equation of L_{AC} is y = \(\frac{1}{8}\) x – 2\(\frac{1}{8}\).

Question 5.

Write the equation for the line that goes through point ( – 10, 8) with slope m = 6.

Answer:

8 = 6( – 10) + b

8 = – 60 + b

68 = b

The equation of the line is y = 6x + 68.

Question 6.

Write the equation for the line that goes through point (12, 15) with slope m = – 2.

Answer:

15 = – 2(12) + b

15 = – 24 + b

39 = b

The equation of the line is y = – 2x + 39.

Question 7.

Write the equation for the line that goes through point (1, 1) with slope m = – 9.

Answer:

1 = – 9(1) + b

1 = – 9 + b

10 = b

The equation of the line is y = – 9x + 10.

Question 8.

Determine the equation of the line that goes through points (1, 1) and (3, 7).

Answer:

The slope of the line is

m = \(\frac{1 – 7}{1 – 3}\)

= \(\frac{ – 6}{ – 2}\)

= 3.

The y – intercept point of the line is

7 = 3(3) + b

7 = 9 + b

– 2 = b.

The equation of the line is y = 3x – 2.

### Eureka Math Grade 8 Module 4 Lesson 21 Exit Ticket Answer Key

Question 1.

Write the equation for the line l shown in the figure below.

Answer:

Using the points ( – 3, 1) and (6, 5), the slope of the line is

m = \(\frac{5 – 1}{6 – ( – 3)}\)

m = \(\frac{4}{9}\)

5 = \(\frac{4}{9}\) (6) + b

5 = \(\frac{8}{3}\) + b

5 – \(\frac{8}{3}\) = \(\frac{8}{3}\) – \(\frac{8}{3}\) + b

\(\frac{7}{3}\) = b

The equation of the line is y = \(\frac{4}{9}\) x + \(\frac{7}{3}\).

Question 2.

A line goes through the point (5, – 7) and has slope m = – 3. Write the equation that represents the line.

Answer:

– 7 = – 3(5) + b

– 7 = – 15 + b

8 = b

The equation of the line is y = – 3x + 8.