# Point-Slope Form 12-4 Warm Up Problem of the Day Lesson Presentation

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Point-Slope Form 12-4 Warm Up Problem of the Day Lesson Presentation
Course 3 Warm Up Problem of the Day Lesson Presentation

Point-Slope Form 12-4 Warm Up
Course 3 12-4 Point-Slope Form Warm Up Write the equation of the line that passes through each pair of points in slope-intercept form. 1. (0, –3) and (2, –3) 2. (5, –3) and (5, 1) 3. (–6, 0) and (0, –2) 4. (4, 6) and (–2, 0) y = –3 x = 5 y = – x – 2 1 3 y = x + 2

Point-Slope Form 12-4 Problem of the Day
Course 3 12-4 Point-Slope Form Problem of the Day Without using equations for horizontal or vertical lines, write the equations of four lines that form a square. Possible answer: y = x + 2, y = x – 2, y = –x + 2, y = –x – 2

Course 3 12-4 Point-Slope Form Learn to find the equation of a line given one point and the slope.

Insert Lesson Title Here
Course 3 12-4 Point-Slope Form Insert Lesson Title Here Vocabulary point-slope form

Course 3 12-4 Point-Slope Form The point-slope form of an equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1). Point on the line Point-slope form y – y1 = m (x – x1) (x1, y1) slope

Course 3 12-4 Point-Slope Form Additional Example 1A: Using Point-Slope Form to Identify Information About a Line Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. y – 7 = 3(x – 4) y – y1 = m(x – x1) The equation is in point-slope form. y – 7 = 3(x – 4) Read the value of m from the equation. m = 3 (x1, y1) = (4, 7) Read the point from the equation. The line defined by y – 7 = 3(x – 4) has slope 3, and passes through the point (4, 7).

Course 3 12-4 Point-Slope Form Additional Example 1B: Using Point-Slope Form to Identify Information About a Line 1 3 y – 1 = (x + 6) y – y1 = m(x – x1) 1 3 y – 1 = (x + 6) y – 1 = [x – (–6)] 1 3 Rewrite using subtraction instead of addition. m = 1 3 (x1, y1) = (–6, 1) The line defined by y – 1 = (x + 6) has slope , and passes through the point (–6, 1). 1 3

Point-Slope Form 12-4 Check It Out: Example 1A
Course 3 12-4 Point-Slope Form Check It Out: Example 1A Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. y – 5 = 2 (x – 2) y – y1 = m(x – x1) The equation is in point-slope form. y – 5 = 2(x – 2) Read the value of m from the equation. m = 2 (x1, y1) = (2, 5) Read the point from the equation. The line defined by y – 5 = 2(x – 2) has slope 2, and passes through the point (2, 5).

Point-Slope Form 12-4 Check It Out: Example 1B 2 3 y – 2 = (x + 3)
Course 3 12-4 Point-Slope Form Check It Out: Example 1B 2 3 y – 2 = (x + 3) y – y1 = m(x – x1) 2 3 y – 2 = (x + 3) y – 2 = [x – (–3)] 2 3 Rewrite using subtraction instead of addition. m = 2 3 (x1, y1) = (–3, 2) The line defined by y – 2 = (x + 3) has slope , and passes through the point (–3, 2). 2 3

Additional Example 2A: Writing the Point-Slope Form of an Equation
Course 3 12-4 Point-Slope Form Additional Example 2A: Writing the Point-Slope Form of an Equation Write the point-slope form of the equation with the given slope that passes through the indicated point. the line with slope 4 passing through (5, –2) y – y1 = m(x – x1) Substitute 5 for x1, –2 for y1, and 4 for m. [y – (–2)] = 4(x – 5) y + 2 = 4(x – 5) The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4(x – 5).

Additional Example 2B: Writing the Point-Slope Form of an Equation
Course 3 12-4 Point-Slope Form Additional Example 2B: Writing the Point-Slope Form of an Equation the line with slope –5 passing through (–3, 7) y – y1 = m(x – x1) Substitute –3 for x1, 7 for y1, and –5 for m. y – 7 = -5[x – (–3)] y – 7 = –5(x + 3) The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5(x + 3).

Point-Slope Form 12-4 Check It Out: Example 2A
Course 3 12-4 Point-Slope Form Check It Out: Example 2A Write the point-slope form of the equation with the given slope that passes through the indicated point. the line with slope 2 passing through (2, –2) y – y1 = m(x – x1) Substitute 2 for x1, –2 for y1, and 2 for m. [y – (–2)] = 2(x – 2) y + 2 = 2(x – 2) The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2(x – 2).

Point-Slope Form 12-4 Check It Out: Example 2B
Course 3 12-4 Point-Slope Form Check It Out: Example 2B the line with slope –4 passing through (–2, 5) y – y1 = m(x – x1) Substitute –2 for x1, 5 for y1, and –4 for m. y – 5 = –4[x – (–2)] y – 5 = –4(x + 2) The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4(x + 2).

Course 3 12-4 Point-Slope Form Additional Example 3: Entertainment Application A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet. As x increases by 30, y increases by 20, so the slope of the line is or . The line passes through the point (0, 18). 20 30 2 3

Course 3 12-4 Point-Slope Form Additional Example 3 Continued y – y1 = m(x – x1) Substitute 0 for x1, 18 for y1, and for m. 2 3 y – 18 = (x – 0) 2 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = x. Substitute 150 for x to find the value of y. 2 3 y – 18 = (150) 2 3 y – 18 = 100 y = 118 The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward.

Point-Slope Form 12-4 Check It Out: Example 3
Course 3 12-4 Point-Slope Form Check It Out: Example 3 A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet. As x increases by 45, y increases by 15, so the slope of the line is or . The line passes through the point (0, 15). 15 45 1 3

Check It Out: Example 3 Continued
Course 3 12-4 Point-Slope Form Check It Out: Example 3 Continued y – y1 = m(x – x1) Substitute 0 for x1, 15 for y1, and for m. 1 3 y – 15 = (x – 0) 1 3 The equation of the line the roller coaster travels along, in point-slope form, is y – 15 = x. Substitute 300 for x to find the value of y. 1 3 y – 15 = (300) 1 3 y – 15 = 100 y = 115 The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward.

Insert Lesson Title Here
Course 3 12-4 Point-Slope Form Insert Lesson Title Here Lesson Quiz Use the point-slope form of each equation to identify a point the line passes through and the slope of the line. 1. y + 6 = 2(x + 5) 2. y – 4 = – (x – 6) Write the point-slope form of the equation with the given slope that passes through the indicated point. 3. the line with slope 4 passing through (3, 5) 4. the line with slope –2 passing through (–2, 4) (–5, –6), 2 2 5 (6, 4), – 2 5 y – 5 = 4(x – 3) y – 4 = –2(x + 2)

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