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**2.4 Writing Equations for Linear Lines**

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**1. 2. Lesson 2.4, For use with pages 98-104**

In a computer generated drawing, a line is represented by the equation 2x – 5y = 15. 1. Solve the equation for y and identify the slope of the line. 5 ANSWER y = 2 x – 3; 2. What should the slope of a second line in the drawing be if that line must be perpendicular to the first line? ANSWER 5 – 2

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EXAMPLE 1 Write an equation given the slope and y-intercept Write an equation of the line shown.

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**Use slope-intercept form.**

EXAMPLE 1 Write an equation given the slope and y-intercept SOLUTION From the graph, you can see that the slope is m = and the y-intercept is b = – 2. Use slope-intercept form to write an equation of the line. 3 4 y = mx + b Use slope-intercept form. y = x + (– 2) 3 4 Substitute for m and –2 for b. 3 4 3 4 y = x (– 2) Simplify.

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**Use slope - intercept form.**

GUIDED PRACTICE for Example 1 Write an equation of the line that has the given slope and y - intercept. 1. m = 3, b = 1 SOLUTION Use slope – intercept point form to write an equation of the line y = mx + b Use slope - intercept form. y = x + 1 3 Substitute 3 for m and 1 for b. y = x + 1 3 Simplify.

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**Use slope - intercept form.**

GUIDED PRACTICE for Example 1 2. m = – 2 , b = – 4 SOLUTION Use slope – intercept point form to write an equation of the line y = mx + b Use slope - intercept form. y = – 2x + (– 4 ) Substitute – 2 for m and – 4 for b. y = – 2x – 4 Simplify.

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**Use slope - intercept form.**

GUIDED PRACTICE for Example 1 3. m = – b = 3 4 7 2 SOLUTION Use slope – intercept point form to write an equation of the line y = mx + b Use slope - intercept form. y = – x + 3 4 7 2 3 4 7 2 Substitute – for m and for b. y = – x + 3 4 7 2 Simplify.

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**Distributive property**

EXAMPLE 2 Write an equation given the slope and a point Write an equation of the line that passes through (5, 4) and has a slope of – 3. SOLUTION Because you know the slope and a point on the line, use point-slope form to write an equation of the line. Let (x1, y1) = (5, 4) and m = – 3. y – y1 = m(x – x1) Use point-slope form. y – 4 = – 3(x – 5) Substitute for m, x1, and y1. y – 4 = – 3x + 15 Distributive property y = – 3x + 19 Write in slope-intercept form.

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EXAMPLE 3 Write equations of parallel or perpendicular lines Write an equation of the line that passes through (–2,3) and is (a) parallel to, and (b) perpendicular to, the line y = –4x + 1. SOLUTION a. The given line has a slope of m1 = –4. So, a line parallel to it has a slope of m2 = m1 = –4. You know the slope and a point on the line, so use the point-slope form with (x1, y1) = (– 2, 3) to write an equation of the line.

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**Distributive property**

EXAMPLE 3 Write equations of parallel or perpendicular lines y – y1 = m2(x – x1) Use point-slope form. y – 3 = – 4(x – (– 2)) Substitute for m2, x1, and y1. y – 3 = – 4(x + 2) Simplify. y – 3 = – 4x – 8 Distributive property y = – 4x – 5 Write in slope-intercept form.

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**Distributive property**

EXAMPLE 3 Write equations of parallel or perpendicular lines b. A line perpendicular to a line with slope m1 = – 4 has a slope of m2 = – = . Use point-slope form with (x1, y1) = (– 2, 3) 1 4 m1 y – y1 = m2(x – x1) Use point-slope form. y – 3 = ( x – (– 2)) 1 4 Substitute for m2, x1, and y1. y – 3 = ( x +2) 1 4 Simplify. y – 3 = x + 1 4 2 Distributive property y = x + 1 4 2 Write in slope-intercept form.

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**Distributive property**

GUIDED PRACTICE GUIDED PRACTICE for Examples 2 and 3 4. Write an equation of the line that passes through (– 1, 6) and has a slope of 4. SOLUTION Because you know the slope and a point on the line, use the point-slope form to write an equation of the line. Let (x1, y1) = (–1, 6) and m = 4 y – y1 = m(x – x1) Use point-slope form. y – 6 = 4(x – (– 1)) Substitute for m, x1, and y1. y – 6 = 4x + 4 Distributive property y = 4x + 10 Write in slope-intercept form.

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**Distributive property Write in slope-intercept form.**

GUIDED PRACTICE GUIDED PRACTICE for Examples 2 and 3 5. Write an equation of the line that passes through (4, –2) and is (a) parallel to, and (b) perpendicular to, the line y = 3x – 1. SOLUTION The given line has a slope of m1 = 3. So, a line parallel to it has a slope of m2 = m1 = 3. You know the slope and a point on the line, so use the point - slope form with (x1, y1) = (4, – 2) to write an equation of the line. y – y1 = m2(x – x1) Use point-slope form. y – (– 2) = 3(x – 4) Substitute for m2, x1, and y1. y + 2 = (x – 4) Simplify. y + 2 = 3x – 12 Distributive property y = 3x – 14 Write in slope-intercept form.

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GUIDED PRACTICE GUIDED PRACTICE for Examples 2 and 3 b. A line perpendicular to a line with slope m1 = 3 has a slope of m2 = – = – 1 1 3 m1 Use point - slope form with (x1, y1) = (4, – 2) y – y1 = m2(x – x1) Use point-slope form. y – (– 2) = – ( x – 4) 1 3 Substitute for m2, x1, and y1. y + 2 = – ( x – 4) 1 3 Simplify. 4 3 y + 2 = – x – 1 Distributive property y = – x – 1 3 2 Write in slope-intercept form.

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EXAMPLE 4 Write an equation given two points Write an equation of the line that passes through (5, –2) and (2, 10). SOLUTION The line passes through (x1, y1) = (5,– 2) and (x2, y2) = (2, 10). Find its slope. y2 – y1 m = x2 –x1 10 – (– 2) = 2 – 5 12 – 3 = – 4

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**Distributive property**

EXAMPLE 4 Write an equation given two points You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. Choose (x1, y1) = (4, – 7). y2 – y1 = m(x – x1) Use point-slope form. y – 10 = – 4(x – 2) Substitute for m, x1, and y1. y – 10 = – 4x + 8 Distributive property y = – 4x + 8 Write in slope-intercept form.

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EXAMPLE 5 Write a model using slope-intercept form Sports In the school year ending in 1993, 2.00 million females participated in U.S. high school sports. By 2003,the number had increased to 2.86 million. Write a linear equation that models female sports participation.

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EXAMPLE 5 Write a model using slope-intercept form SOLUTION STEP 1 Define the variables. Let x represent the time (in years) since 1993 and let y represent the number of participants (in millions). STEP 2 Identify the initial value and rate of change. The initial value is The rate of change is the slope m.

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**Use (x1, y1) = (0, 2.00) and (x2, y2) = (10, 2.86). EXAMPLE 5**

Write a model using slope-intercept form y2 – y1 m = x2 –x1 2.86 – 2.00 = 10 – 0 Use (x1, y1) = (0, 2.00) and (x2, y2) = (10, 2.86). 0.86 = 10 = 0.086 STEP 3 Write a verbal model. Then write a linear equation.

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EXAMPLE 5 Write a model using slope-intercept form y = x ANSWER In slope-intercept form, a linear model is y = 0.086x

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GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 Write an equation of the line that passes through the given points. 6. (– 2, 5), (4, – 7) SOLUTION The line passes through (x1, y1) = (– 2, 5) and (x2, y2) = (4, – 7). Find its slope. y2 – y1 m = x2 –x1 – 7 – 5 = 4 – (– 2) = – 2

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**Distributive property**

GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. Choose (x1, y1) = (4, – 7). y – y1 = m(x – x1) Use point-slope form. y – 7 = – 2(x – 4) Substitute for m, x1, and y1. Simplify y – 7 = – 2 (x + 4) y + 7 = – 2x + 8 Distributive property y = – 2x + 1 Write in slope-intercept form.

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GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 7. (6, 1), (–3, –8) SOLUTION The line passes through (x1, y1) = (6, 1) and (x2, y2) = (–3, –8). Find its slope. y2 – y1 m = x2 –x1 – 8 – 1 = – 3 – 6 – 9 = 1

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**Distributive property**

GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. Choose (x1, y1) = (– 3, – 8). y – y1 = m(x – x1) Use point-slope form. y – (– 8)) = 1(x – (– 3)) Substitute for m, x, and y1. y + 8 = 1 (x + 3) Simplify y + 8 = x + 3 Distributive property y = x – 5 Write in slope-intercept form.

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GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 8. (–1, 2), (10, 0) SOLUTION The line passes through (x1, y1) = (– 1, 2) and (x2, y2) = (10, 0). Find its slope. y2 – y1 m = x2 –x1 0 – 2 = 10– (– 1) 2 11 –

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**Distributive property**

GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 You know the slope and a point on the line, so use point-slope form with either given point to write an equation of the line. Choose (x1, y1) = (10, 0). y – y1 = m(x – x1) Use point-slope form. y – 0 = (x – 10 ) 2 11 – Substitute for m, x, and y1. y = (x – 10) 2 11 – Simplify y = 2 11 – x + 20 Distributive property 2 11 – x + 20 = Write in slope-intercept form.

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GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 9. Sports In Example 5, the corresponding data for males are 3.42 million participants in 1993 and 3.99 million participants in Write a linear equation that models male participation in U.S. high school sports.

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GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 SOLUTION STEP 1 Define the variables. Let x represent the time (in years) since 1993 and let y represent the number of participants (in millions). STEP 2 Identify the initial value and rate of change. The initial value is The rate of change is the slope m.

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**Use (x1, y1) = 3.42 and (x2, y2) = 3.99 GUIDED PRACTICE**

for Examples 4 and 5 y2 – y1 m = x2 –x1 3.99 – 3.42 = 10 – 0 Use (x1, y1) = 3.42 and (x2, y2) = 3.99 = STEP 3 Write a verbal model. Then write a linear equation.

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GUIDED PRACTICE GUIDED PRACTICE for Examples 4 and 5 y = x ANSWER In slope-intercept form, a linear model is y = 0.057x

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EXAMPLE 6 Write a model using standard form Online Music You have $30 to spend on downloading songs for your digital music player. Company A charges $.79 per song, and company B charges $.99 per song. Write an equation that models this situation. SOLUTION Write a verbal model. Then write an equation.

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EXAMPLE 6 Write a model using standard form x y = ANSWER An equation for this situation is 0.79x y = 30.

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GUIDED PRACTICE for Example 6 What If? In Example 6, suppose that company A charges $.69 per song and company B charges $.89 per song. Write an equation that models this situation.

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GUIDED PRACTICE for Example 6 SOLUTION x y = ANSWER An equation for this situation is .69x + .89y = 30.

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