{"id":85837,"date":"2022-05-30T11:10:39","date_gmt":"2022-05-30T05:40:39","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=85837"},"modified":"2022-05-10T14:19:54","modified_gmt":"2022-05-10T08:49:54","slug":"eureka-math-grade-8-module-4-lesson-21","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-4-lesson-21\/","title":{"rendered":"Eureka Math Grade 8 Module 4 Lesson 21 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 4 Lesson 21 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 4 Lesson 21 Example Answer Key<\/h3>\n

Example 1.
\nLet a line l be given in the coordinate plane. What linear equation is the graph of line l?
\n\"Engage
\nAnswer:
\nWe can pick two points to determine the slope, but the precise location of the y – intercept point cannot be determined from the graph.
\nCalculate the slope of the line.
\nUsing points ( – 2, 2) and (5, 4), the slope of the line is
\nm = \\(\\frac{2 – 4}{ – 2 – 5}\\)
\n= \\(\\frac{ – 2}{ – 7}\\)
\n= \\(\\frac{2}{7}\\)
\n\u2192 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.<\/p>\n

\u2192 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.
\nYes. We can substitute one of the points and the slope into the equation and solve for b.<\/p>\n

\u2192 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)?
\nNo, because there can be only one line with a given slope that goes through a point.<\/p>\n

\u2192 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.
\nSample student work:
\n2 = \\(\\frac{2}{7}\\) ( – 2) + b
\n2 = – \\(\\frac{4}{7}\\) + b
\n2 + \\(\\frac{4}{7}\\) = – \\(\\frac{4}{7}\\) + \\(\\frac{4}{7}\\) + b
\n\\(\\frac{18}{7}\\) = b<\/p>\n

4 = \\(\\frac{2}{7}\\) (5) + b
\n4 = \\(\\frac{10}{7}\\) + b
\n4 – \\(\\frac{10}{7}\\) = \\(\\frac{10}{7}\\) – \\(\\frac{10}{7}\\) + b
\n\\(\\frac{18}{7}\\) = b
\nThe y – intercept point is at (0, \\(\\frac{18}{7}\\)), and the equation of the line is y = \\(\\frac{2}{7}\\) x + \\(\\frac{18}{7}\\).<\/p>\n

\u2192 The equation of the line is
\ny = \\(\\frac{2}{7}\\) x + \\(\\frac{18}{7}\\).
\n\u2192 Write it in standard form.
\nSample student work:
\n(y = \\(\\frac{2}{7}\\) x + \\(\\frac{18}{7}\\))7
\n7y = 2x + 18
\n– 2x + 7y = 2x – 2x + 18
\n– 2x + 7y = 18
\n– 1( – 2x + 7y = 18)
\n2x – 7y = – 18<\/p>\n

Example 2.
\nLet a line l be given in the coordinate plane. What linear equation is the graph of line l?
\n\"Engage
\nAnswer:
\nDetermine the slope of the line.
\nUsing points ( – 1, 4) and (4, 1), the slope of the line is
\nm = \\(\\frac{4 – 1}{ – 1 – 4}\\)
\n= \\(\\frac{3}{ – 5}\\)
\n= – \\(\\frac{3}{5}\\).<\/p>\n

Determine the y – intercept point of the line.
\nSample student work:
\n4 = ( – \\(\\frac{3}{5}\\))( – 1) + b
\n4 = \\(\\frac{3}{5}\\) + b
\n4 – \\(\\frac{3}{5}\\) = \\(\\frac{3}{5}\\) – \\(\\frac{3}{5}\\) + b
\n\\(\\frac{17}{5}\\) = b
\nThe y – intercept point is at (0, \\(\\frac{17}{5}\\)).<\/p>\n

\u2192 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.
\ny = – \\(\\frac{3}{5}\\) x + \\(\\frac{17}{5}\\)
\n\u2192 Transform the equation so that it is written in standard form.
\nSample student work:
\ny = – \\(\\frac{3}{5}\\) x + \\(\\frac{17}{5}\\)
\n(y = – \\(\\frac{3}{5}\\) x + \\(\\frac{17}{5}\\))5
\n5y = – 3x + 17
\n3x + 5y = – 3x + 3x + 17
\n3x + 5y = 17<\/p>\n

Example 3.
\nLet a line l be given in the coordinate plane. What linear equation is the graph of line l?
\n\"Engage
\nAnswer:
\n\u2192 Using points (12, 2) and (13, 7), the slope of the line is
\nm = \\(\\frac{2 – 7}{12 – 13}\\)
\n= \\(\\frac{ – 5}{ – 1}\\)
\n= 5.<\/p>\n

\u2192 Now, determine the y – intercept point of the line, and write the equation of the line in slope – intercept form.
\nSample student work:
\n2 = 5(12) + b
\n2 = 60 + b
\nb = – 58
\nThe y – intercept point is at (0, – 58), and the equation of the line is y = 5x – 58.<\/p>\n

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.
\nSample student work:
\ny = 5x – 58
\n– 5x + y = 5x – 5x – 58
\n– 5x + y = – 58
\n– 1( – 5x + y = – 58)
\n5x – y = 58<\/p>\n

Example 4.
\nLet a line l be given in the coordinate plane. What linear equation is the graph of line l?
\n\"Engage
\nAnswer:
\nUsing points (3, 1) and ( – 3, – 1), the slope of the line is
\nm = \\(\\frac{ – 1 – 1}{ – 3 – 3}\\)
\n= \\(\\frac{ – 2}{ – 6}\\)
\n= \\(\\frac{1}{3}\\)
\nThe y – intercept point is at (0, 0), and the equation of the line is y = \\(\\frac{1}{3}\\) x.<\/p>\n

Eureka Math Grade 8 Module 4 Lesson 21 Exercise Answer Key<\/h3>\n

Exercises<\/p>\n

Exercise 1.
\nWrite the equation for the line l shown in the figure.
\n\"Engage
\nAnswer:
\nUsing the points ( – 1, – 3) and (2, – 2), the slope of the line is
\nm = \\(\\frac{ – 3 – ( – 2)}{ – 1 – 2}\\)
\n= \\(\\frac{ – 1}{ – 3}\\)
\n= \\(\\frac{1}{3}\\) .
\n– 2 = \\(\\frac{1}{3}\\) (2) + b
\n– 2 = \\(\\frac{2}{3}\\) + b
\n– 2 – \\(\\frac{2}{3}\\) = \\(\\frac{2}{3}\\) – \\(\\frac{2}{3}\\) + b
\n– \\(\\frac{8}{3}\\) = b
\nThe equation of the line is y = \\(\\frac{1}{3}\\) x – \\(\\frac{8}{3}\\).<\/p>\n

Exercise 2.
\nWrite the equation for the line l shown in the figure.
\n\"Engage
\nAnswer:
\nUsing the points ( – 3, 7) and (2, 8), the slope of the line is
\nm = \\(\\frac{7 – 8}{ – 3 – 2}\\)
\n= \\(\\frac{ – 1}{ – 5}\\)
\n= \\(\\frac{1}{5}\\).
\n8 = \\(\\frac{1}{5}\\) (2) + b
\n8 = \\(\\frac{2}{5}\\) + b
\n8 – \\(\\frac{2}{5}\\) = \\(\\frac{2}{5}\\) – \\(\\frac{2}{5}\\) + b
\n\\(\\frac{38}{5}\\) = b
\nThe equation of the line is y = \\(\\frac{1}{5}\\) x + \\(\\frac{38}{5}\\).<\/p>\n

Exercise 3.
\nDetermine the equation of the line that goes through points ( – 4, 5) and (2, 3).
\nAnswer:
\nThe slope of the line is
\nm = \\(\\frac{5 – 3}{ – 4 – 2}\\)
\n= \\(\\frac{2}{ – 6}\\)
\n= – \\(\\frac{1}{3}\\)
\nThe y – intercept point of the line is
\n3 = – \\(\\frac{1}{3}\\) (2) + b
\n3 = – \\(\\frac{2}{3}\\) + b
\n\\(\\frac{11}{3}\\) = b.
\nThe equation of the line is y = – \\(\\frac{1}{3}\\) x + \\(\\frac{11}{3}\\).<\/p>\n

Exercise 4.
\nWrite the equation for the line l shown in the figure.
\n\"Engage
\nAnswer:
\nUsing the points ( – 7, 2) and ( – 6, – 2), the slope of the line is
\nm = \\(\\frac{2 – ( – 2)}{ – 7 – ( – 6)}\\)
\n= \\(\\frac{4}{ – 1}\\)
\n= – 4.
\n– 2 = – 4( – 6) + b
\n– 2 = 24 + b
\n– 26 = b
\nThe equation of the line is y = – 4x – 26.<\/p>\n

Exercise 5.
\nA line goes through the point (8, 3) and has slope m = 4. Write the equation that represents the line.
\nAnswer:
\n3 = 4(8) + b
\n3 = 32 + b
\n– 29 = b
\nThe equation of the line is y = 4x – 29.<\/p>\n

Eureka Math Grade 8 Module 4 Lesson 21 Problem Set Answer Key<\/h3>\n

Question 1.
\nWrite the equation for the line l shown in the figure.
\n\"Eureka
\nAnswer:
\nUsing the points ( – 3, 2) and (2, – 2), the slope of the line is
\nm = \\(\\frac{2 – ( – 2)}{ – 3 – 2}\\)
\n= \\(\\frac{4}{ – 5}\\)
\n= – \\(\\frac{4}{5}\\)
\n2 = ( – \\(\\frac{4}{5}\\))( – 3) + b
\n2 = \\(\\frac{12}{5}\\) + b
\n2 – \\(\\frac{12}{5}\\) = \\(\\frac{12}{5}\\) – \\(\\frac{12}{5}\\) + b
\n– \\(\\frac{2}{5}\\) = b
\nThe equation of the line is y = – \\(\\frac{4}{5}\\) x – \\(\\frac{2}{5}\\).<\/p>\n

Question 2.
\nWrite the equation for the line l shown in the figure.
\n\"Eureka
\nAnswer:
\nUsing the points ( – 6, 2) and ( – 5, 5), the slope of the line is
\nm = \\(\\frac{2 – 5}{ – 6 – ( – 5)}\\)
\n= \\(\\frac{ – 3}{ – 1}\\)
\n= 3.
\n5 = 3( – 5) + b
\n5 = – 15 + b
\n20 = b
\nThe equation of the line is y = 3x + 20.<\/p>\n

Question 3.
\nWrite the equation for the line l shown in the figure.
\n\"Eureka
\nAnswer:
\nUsing the points ( – 3, 1) and (2, 2), the slope of the line is
\nm = \\(\\frac{1 – 2}{ – 3 – 2}\\)
\n= \\(\\frac{ – 1}{ – 5}\\)
\n= \\(\\frac{1}{5}\\)
\n2 = \\(\\frac{1}{5}\\) (2) + b
\n2 = \\(\\frac{2}{5}\\) + b
\n2 – \\(\\frac{2}{5}\\) = \\(\\frac{2}{5}\\) – \\(\\frac{2}{5}\\) + b
\n\\(\\frac{8}{5}\\) = b
\nThe equation of the line is y = \\(\\frac{1}{5}\\) x + \\(\\frac{8}{5}\\).<\/p>\n

Question 4.
\nTriangle ABC is made up of line segments formed from the intersection of lines LAB<\/sub>, LBC<\/sub>, and LAC<\/sub>. Write the equations that represent the lines that make up the triangle.
\n\"Eureka
\nAnswer:
\nA( – 3, – 3), B(3, 2), C(5, – 2)
\nThe slope of LAB<\/sub>:
\nm = \\(\\frac{ – 3 – 2}{ – 3 – 3}\\)
\n= \\(\\frac{ – 5}{ – 6}\\)
\n= \\(\\frac{5}{6}\\)
\n2 = \\(\\frac{5}{6}\\) (3) + b
\n2 = \\(\\frac{5}{2}\\) + b
\n2 – \\(\\frac{5}{2}\\) = \\(\\frac{5}{2}\\) – \\(\\frac{5}{2}\\) + b
\n– \\(\\frac{1}{2}\\) = b
\nThe equation of LAB<\/sub> is y = \\(\\frac{5}{6}\\) x – \\(\\frac{1}{2}\\).<\/p>\n

The slope of LBC<\/sub>:
\nm = \\( = \\frac{2 – ( – 2)}{3 – 5}\\)
\n= \\(\\frac{4}{ – 2}\\)
\n= – 2
\n2 = – 2(3) + b
\n2 = – 6 + b
\n8 = b
\nThe equation of LBC<\/sub> is y = – 2x + 8.<\/p>\n

The slope of LAC<\/sub>:
\nm = \\(\\frac{ – 3 – ( – 2)}{ – 3 – 5}\\)
\n= \\(\\frac{ – 1}{ – 8}\\)
\n= \\(\\frac{1}{8}\\)
\n– 2 = \\(\\frac{1}{8}\\) (5) + b
\n– 2 = \\(\\frac{5}{8}\\) + b
\n– 2 – \\(\\frac{5}{8}\\) = \\(\\frac{5}{8}\\) – \\(\\frac{5}{8}\\) + b
\n– 2\\(\\frac{1}{8}\\) = b
\nThe equation of LAC<\/sub> is y = \\(\\frac{1}{8}\\) x – 2\\(\\frac{1}{8}\\).<\/p>\n

Question 5.
\nWrite the equation for the line that goes through point ( – 10, 8) with slope m = 6.
\nAnswer:
\n8 = 6( – 10) + b
\n8 = – 60 + b
\n68 = b
\nThe equation of the line is y = 6x + 68.<\/p>\n

Question 6.
\nWrite the equation for the line that goes through point (12, 15) with slope m = – 2.
\nAnswer:
\n15 = – 2(12) + b
\n15 = – 24 + b
\n39 = b
\nThe equation of the line is y = – 2x + 39.<\/p>\n

Question 7.
\nWrite the equation for the line that goes through point (1, 1) with slope m = – 9.
\nAnswer:
\n1 = – 9(1) + b
\n1 = – 9 + b
\n10 = b
\nThe equation of the line is y = – 9x + 10.<\/p>\n

Question 8.
\nDetermine the equation of the line that goes through points (1, 1) and (3, 7).
\nAnswer:
\nThe slope of the line is
\nm = \\(\\frac{1 – 7}{1 – 3}\\)
\n= \\(\\frac{ – 6}{ – 2}\\)
\n= 3.
\nThe y – intercept point of the line is
\n7 = 3(3) + b
\n7 = 9 + b
\n– 2 = b.
\nThe equation of the line is y = 3x – 2.<\/p>\n

Eureka Math Grade 8 Module 4 Lesson 21 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nWrite the equation for the line l shown in the figure below.
\n\"Eureka
\nAnswer:
\nUsing the points ( – 3, 1) and (6, 5), the slope of the line is
\nm = \\(\\frac{5 – 1}{6 – ( – 3)}\\)
\nm = \\(\\frac{4}{9}\\)
\n5 = \\(\\frac{4}{9}\\) (6) + b
\n5 = \\(\\frac{8}{3}\\) + b
\n5 – \\(\\frac{8}{3}\\) = \\(\\frac{8}{3}\\) – \\(\\frac{8}{3}\\) + b
\n\\(\\frac{7}{3}\\) = b
\nThe equation of the line is y = \\(\\frac{4}{9}\\) x + \\(\\frac{7}{3}\\).<\/p>\n

Question 2.
\nA line goes through the point (5, – 7) and has slope m = – 3. Write the equation that represents the line.
\nAnswer:
\n– 7 = – 3(5) + b
\n– 7 = – 15 + b
\n8 = b
\nThe equation of the line is y = – 3x + 8.<\/p>\n","protected":false},"excerpt":{"rendered":"

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 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85837"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=85837"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85837\/revisions"}],"predecessor-version":[{"id":137171,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/85837\/revisions\/137171"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=85837"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=85837"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=85837"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}