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Points on a Line Topic 4.2.1

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**Points on a Line 4.2.1 Topic California Standards:**

6.0 Students graph a linear equation and compute the x- and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 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

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Topic 4.2.1 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. (1, 1) (2, 3) (3, 5) (–1, –3)

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Topic 4.2.1 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: Linear: 3x + y = 4, x = 6, y = 5 – x Nonlinear: xy = 12, x2 + 3y = 1, y3 = 20

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Topic 4.2.1 Points on a Line Linear equations in two variables, x and y, can be written in the form: Ax + By = C The solution set to the equation Ax + By = C consists of all ordered pairs (x, y) that satisfy the equation. All the points in this solution set lie on a straight line. This straight line is the graph of the equation.

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Topic 4.2.1 Points on a Line If the ordered pair (x, y) satisfies the equation Ax + By = C, then the point (x, y) lies on the graph of the equation. The ordered pair (–2, 5) satisfies the equation 2x + 2y = 6. So the point (–2, 5) lies on the line 2x + 2y = 6. The point (2, 1) lies on the line 2x + 2y = 6. So the ordered pair (2, 1) satisfies the equation 2x + 2y = 6.

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**Points on a Line 4.2.1 Topic 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.

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Topic 4.2.1 Points on a Line Example 1 a) Show that the point (2, –3) lies on the graph of x – 3y = 11. b) Determine whether the point (–1, 1) lies on the graph of 2x + 3y = 4. Solution a) 2 – 3(–3) = 11 Substitute 2 for x and –3 for y 2 + 9 = 11 11 = 11 A true statement So the point (2, –3) lies on the graph of x – 3y = 11, since (2, –3) satisfies the equation x – 3y = 11. Solution continues… Solution follows…

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Topic 4.2.1 Points on a Line Example 1 a) Show that the point (2, –3) lies on the graph of x – 3y = 11. b) Determine whether the point (–1, 1) lies on the graph of 2x + 3y = 4. Solution (continued) b) If (–1, 1) lies on the graph of 2x + 3y = 4, then 2(–1) + 3(1) = 4. But 2(–1) + 3(1) = –2 + 3 = 1 Since 1 ¹ 4, (–1, 1) does not lie on the graph of 2x + 3y = 4.

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**Points on a Line 4.2.1 Topic Guided Practice**

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

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**Points on a Line 4.2.1 Topic Guided Practice**

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

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**Points on a Line 4.2.1 Topic Guided Practice**

Determine whether or not each point lies on the line of the given equation. (1, 4); 4y – 12x = 3 ( , – ); 6x – 16y = 7 ( , – ); –3x – 10y = 2 no: 4(4) – 12(1) = 4 ¹ 3 1 3 1 4 no: 6( ) – 16(– ) = 6 ¹ 7 1 3 4 2 3 2 5 yes: –3( ) – 10(– ) = 2 2 3 5 Solution follows…

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**Points on a Line 4.2.1 Topic Independent Practice**

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

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**Points on a Line 4.2.1 Topic Independent Practice**

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

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**Points on a Line 4.2.1 Topic Independent Practice**

In Exercises 9–12, determine whether or not each point lies on the graph of 6x – 6y = 24. (4, 0) (1, –3) (100, 96) (–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 Solution follows…

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**Points on a Line 4.2.1 Topic Independent Practice**

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. 14. Determine whether the point (3, 4) lies on the lines 4x + 6y = 36 and 8x – 7y = 30. 4x + 6y = 36: 4(3) + 6(4) = 36 8x – 7y = 30: 8(3) – 7(4) = –4 ¹ 36 (3, 4) does lie on the line 4x + 6y = 36, but not on the line 8x – 7y = 30, so it doesn’t lie on both lines. Solution follows…

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**Points on a Line 4.2.1 Topic 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.

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Section 1.8 Homework questions?. Section Concepts 1.8 Linear Equations in Two Variables Slide 2 Copyright (c) The McGraw-Hill Companies, Inc. Permission.

Section 1.8 Homework questions?. Section Concepts 1.8 Linear Equations in Two Variables Slide 2 Copyright (c) The McGraw-Hill Companies, Inc. Permission.

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