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

2. 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

3. 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

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

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

6. 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

7. 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).

8. 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

9. 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).

10. 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

11. 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).

12. 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).

13. 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).

14. 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).

15. Additional Example 3: Entertainment Application
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

16. Additional Example 3 Continued
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.

17. 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

18. 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.

19. 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)