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1 Topic 4.2.1 Points on a Line. 2 California Standards: 6.0 Students graph a linear equation and compute the x - and y -intercepts (e.g., graph 2 x +

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Presentation on theme: "1 Topic 4.2.1 Points on a Line. 2 California Standards: 6.0 Students graph a linear equation and compute the x - and y -intercepts (e.g., graph 2 x +"— Presentation transcript:

1 1 Topic Points on a Line

2 2 California Standards: 6.0 Students graph a linear equation and compute the x - and y -intercepts (e.g., graph 2 x + 6 y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2 x + 6 y < 4). 7.0 Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. What it means for you: You’ll learn how to show mathematically that points lie on a line. Key Words: linear equation variable solution set verify Topic 4.2.1

3 3 Points on a Line You already dealt with lines in Topics and In this Topic you’ll see a formal definition relating ordered pairs to a line — and you’ll also learn how to show that points lie on a particular line. Topic (1, 1) (2, 3) (3, 5) (–1, –3)

4 4 Points on a Line Graphs of Linear Equations are Straight Lines An equation is linear if the variables have an exponent of one and there are no variables multiplied together. For example: Topic Linear: 3 x + y = 4, 2 x = 6, y = 5 – x Nonlinear: xy = 12, x y = 1, 8 y 3 = 20

5 5 Points on a Line Linear equations in two variables, x and y, can be written in the form: The solution set to the equation A x + B y = C consists of all ordered pairs ( x, y ) that satisfy the equation. Topic A x + B y = C All the points in this solution set lie on a straight line. This straight line is the graph of the equation.

6 6 Points on a Line Topic If the ordered pair ( x, y ) satisfies the equation A x + B y = C, then the point ( x, y ) lies on the graph of the equation. The ordered pair (–2, 5) satisfies the equation 2 x + 2 y = 6. So the point (–2, 5) lies on the line 2 x + 2 y = 6. The point (2, 1) lies on the line 2 x + 2 y = 6. So the ordered pair (2, 1) satisfies the equation 2 x + 2 y = 6.

7 7 Points on a Line Verifying That Points Lie on a Line To determine whether a point ( x, y ) lies on the line of a given equation, you need to find out whether the ordered pair ( x, y ) satisfies the equation. If it does, the point is on the line. You do this by substituting x and y into the equation. Topic 4.2.1

8 8 Solution continues… Points on a Line Example 1 a) Show that the point (2, –3) lies on the graph of x – 3 y = 11. Solution follows… Topic Solution 11 = 11 So the point (2, –3) lies on the graph of x – 3 y = 11, since (2, –3) satisfies the equation x – 3 y = 11. a) 2 – 3(–3) = 11Substitute 2 for x and –3 for y = 11 A true statement b) Determine whether the point (–1, 1) lies on the graph of 2 x + 3 y = 4.

9 9 Points on a Line Example 1 a) Show that the point (2, –3) lies on the graph of x – 3 y = 11. Topic Solution (continued) b) Determine whether the point (–1, 1) lies on the graph of 2 x + 3 y = 4. But 2(–1) + 3(1) = –2 + 3 Since 1  4, (–1, 1) does not lie on the graph of 2 x + 3 y = 4. b) If (–1, 1) lies on the graph of 2 x + 3 y = 4, then 2(–1) + 3(1) = 4. = 1

10 10 Points on a Line Guided Practice Solution follows… 1. (–1, 2); 2 x – y = –4 2. (3, –4); –2 x – 3 y = 6 3. (–3, –1); –5 x + 3 y = (–7, –3); 2 y – 3 x = 15 Determine whether or not each point lies on the line of the given equation. yes: 2(–1) – 2 = –4 Topic no: –5(–3) + 3(–1) = –18  11 yes: –2(3) – 3(–4) = 6 yes: 2(–3) – 3(–7) = 15

11 11 Points on a Line Guided Practice Solution follows… 5. (–2, –2); y = 3 x (–5, –3); – y + 2 x = –7 7. (–2, –1); 8 x – 15 y = 3 Determine whether or not each point lies on the line of the given equation. Topic yes: –2 = 3(–2) + 4 no: 8(–2) – 15(–1) = –1  3 yes: –(–3) + 2(–5) = –7

12 12 Points on a Line Guided Practice Solution follows… Determine whether or not each point lies on the line of the given equation. Topic (1, 4); 4 y – 12 x = 3 9. (, – );6 x – 16 y = (, – );–3 x – 10 y = 2 no: 4(4) – 12(1) = 4  no: 6( ) – 16(– ) = 6  yes: –3( ) – 10(– ) =

13 13 Independent Practice Solution follows… In Exercises 1–4, determine whether or not each point lies on the graph of 5 x – 4 y = 20. Points on a Line Topic (0, 4) 2. (4, 0) 3. (2, –3) 4. (8, 5) no: 5(0) – 4(4) = –16  20 yes: 5(4) – 4(0) = 20 no: 5(2) – 4(–3) = 22  20 yes: 5(8) – 4(5) = 20

14 14 Independent Practice Solution follows… In Exercises 5–8, determine whether or not each point lies on the graph of 6 x + 3 y = 15. Points on a Line Topic (2, 1) 6. (0, 5) 7. (–1, 6) 8. (3, –1) yes: 6(2) + 3(1) = 15 yes: 6(0) + 3(5) = 15 no: 6(–1) + 3(6) = 12  15 yes: 6(3) + 3(–1) = 15

15 15 Independent Practice Solution follows… In Exercises 9–12, determine whether or not each point lies on the graph of 6 x – 6 y = 24. Points on a Line Topic (4, 0) 10. (1, –3) 11. (100, 96) 12. (–400, –404) yes: 6(4) – 6(0) = 24 yes: 6(1) – 6(–3) = 24 yes: 6(100) – 6(96) = 24 yes: 6(–400) – 6(–404) = 24

16 16 Independent Practice Solution follows… 13.Explain in words why (2, 31) is a point on the line x = 2, but not a point on the line y = 2. x = 2 includes all points whose x - coordinate is 2, so (2, 31) is a point on x = 2; the y - coordinate is 31, so (2, 31) does not lie on the line y = 2. 4 x + 6 y = 36: 4(3) + 6(4) = 36 8 x – 7 y = 30: 8(3) – 7(4) = –4  36 (3, 4) does lie on the line 4 x + 6 y = 36, but not on the line 8 x – 7 y = 30, so it doesn’t lie on both lines. Points on a Line Topic Determine whether the point (3, 4) lies on the lines 4 x + 6 y = 36 and 8 x – 7 y = 30.

17 17 Round Up You can always substitute x and y into the equation to prove whether a coordinate pair lies on a line. That’s because if the coordinate pair lies on the line then it’s actually a solution of the equation. Points on a Line Topic 4.2.1


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