Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Find the rate of change of the data in each table. 1. 2. 3. Simplify each expression. 4. –3(x – 5) 5. 5(x + 2) 6. – (x – 6) 4 9 2 5 8 11 –2 –8 –14 x y –3 –5 –1 1 3 –4 10 7.5 2.5 –6 –11 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 1. Use points (2, 4) and (5, –2). rate of change = = = –2 2. Use points (–3, –5) and (–1, –4). rate of change = = = = 3. Use points (10, 4) and (7.5, –1). rate of change = = = = 2 4. –3(x – 5) = –3x – (–3)(5) = –3x + 15 5. 5(x + 2) = 5x + 5(2) = 5x + 10 6. – (x – 6) = – x – (– )(6) = – x + 4 9 8 3 –5 + 4 –3 + 1 4 – (–2) 2 – 5 4 – (–1 ) 10 – 7.5 –5 – (–4) –3 – (–1) 4 + 1 2.5 5 –1 –2 6 –3 1 2 Solutions 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 1 3 Graph the equation y – 2 = (x – 1). 1 3 The equation of a line that passes through (1, 2) with slope . Start at (1, 2). Using the slope, go up 1 unit and right 3 units to (4, 3). Draw a line through the two points. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Write the equation of the line with slope –2 that passes through the point (3, –3). y – y1 = m(x – x1) Substitute (3, –3) for (x1, y1) and –2 for m. y – (–3) = –2(x – 3) Simplify the grouping symbols. y + 3 = –2(x – 3) 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Write equations for the line in point-slope form and in slope-intercept form. The slope is – . 1 3 Step 1  Find the slope. = m y2 – y1 x2 – x1 4 – 3 –1 – 2 = – 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (continued) Step 2  Use either point to write the the equation in point-slope form. Use (–1, 4). y – y1 = m(x – x1) y – 4 = – (x + 1) 1 3 Step 3  Rewrite the equation from Step 2 in slope– intercept form. y – 4 = – (x + 1) y – 4 = – x – y = – x + 3 1 3 2 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Is the relationship shown by the data linear? If so, model the data with an equation. Step 1  Find the rate of change for consecutive ordered pairs. –2 –1 –6 –3 = 2 3 6 2 –1 –3 4 –2 –6 x y –1( ) –2 –3( ) –6 –2( ) –4 The relationship is linear. The rate of change is 2. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 (continued) Step 2 Use the slope 2 and a point (2,4) to write an equation. y – y1 = m(x – x1) Use the point-slope form. y – 4 = 2(x – 2) Substitute (2, 4) for (x1, y1) and 2 for m. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 Is the relationship shown by the data linear? If so, model the data with an equation. Step 1  Find the rate of change for consecutive ordered pairs. 1 2 = 1 – / –2 –1 1 2 x y 1 ( ) 1 2 ( ) 1 The relationship is not linear. 6-4

Point-Slope Form and Writing Linear Equations ALGEBRA 1 LESSON 6-4 1. Graph the equation y + 1 = –(x – 3). 2. Write an equation of the line with slope – that passes through the point (0, 4). 3. Write an equation for the line that passes through (3, –5) and (–2, 1) in Point-Slope form and Slope-Intercept form. 4. Is the relationship shown by the data linear? If so, model that data with an equation. 2 3 y – 4 = – (x – 0), or y = – x + 4 2 3 –10 –7 5 20 –3 –1 x y 6 5 7 y + 5 = – (x – 3); y = – x – 2 5 yes; y + 3 = (x – 0) 6-4