Identify other points on the line by completing the following table.
x: 1, 2, 5, 7, 9, 12, 20
y: 3, __, __, 15, __, __, __
Answer: For y = 2x + 1: x = 1, 2, 5, 7, 9, 12, 20 give y = 3, 5, 11, 15, 19, 25, 41. So (2, 5), (5, 11), (9, 19), (12, 25) and (20, 41) are also on the line.
Step-by-step solution
Given: The line is y = 2x + 1 (it passes through A (0, 1) and B (3, 7))
Idea: A point lies on the line exactly when its coordinates satisfy the equation. So for each x, the matching y is 2x + 1.
- Use y = 2x + 1: x = 2 → y = 5; x = 5 → y = 11.1 mark
- x = 9 → y = 19; x = 12 → y = 25; x = 20 → y = 41.1 mark
- Completed table:½ mark
x 1 2 5 7 9 12 20 y 3 5 11 15 19 25 41 - Plotting the points on graph paper (see the graph), they all lie on the straight line through A (0, 1) and B (3, 7). For example, (7, 15) is on the line because 2 × 7 + 1 = 15.½ mark
The missing values are y = 5, 11, 19, 25 and 41, giving the points (2, 5), (5, 11), (9, 19), (12, 25) and (20, 41) on y = 2x + 1.
Check: Each y is one more than twice x: 2 × 12 + 1 = 25 ✓, 2 × 20 + 1 = 41 ✓. Also, each step of 1 in x adds 2 to y (3 → 5).
Answer to write in the exam
y = 2x + 1
x = 2: y = 5; x = 5: y = 11; x = 9: y = 19
x = 12: y = 25; x = 20: y = 41
∴ y values: 3, 5, 11, 15, 19, 25, 41
Common mistakes that cost marks
- Working out 2x + 1 as 2(x + 1): for x = 5 that gives 12 instead of 11.
- Adding 2 for each column of the table even when the x values jump (2 to 5 is 3 steps, so y rises by 6, not 2).
- Plotting (5, 11) as (11, 5). The x-coordinate comes first.
How this can come in the exam
MCQ (1 mark)
Which point lies on the line y = 2x + 1?
- (4, 8)
- (6, 13)
- (3, 6)
- (10, 20)
Show answer
(B) (6, 13)
2 × 6 + 1 = 13, so (6, 13) satisfies the equation.
Short answer (2 marks)
Complete the table for y = 3x − 2 for x = 0, 1, 4, 10, and check whether (5, 12) lies on the line.
Show answer
y = −2, 1, 10, 28 (1 mark). 3 × 5 − 2 = 13 ≠ 12, so (5, 12) is not on the line (1 mark).Try one yourself
For the line y = 2x + 1, find y when x = −3 and when x = 15.
Show answer
−5 and 31, giving the points (−3, −5) and (15, 31).
More questions like this
- Let us plot the points (−1, −3), (0, 0), (1, 3), (3, 9), (4, 12) in the coordinate plane on a graph paper as shown in the figure. Join the points (−1, −3) and (4, 12) using a ruler. Doing so, observe that all five points lie on a straight line. Can you guess the equation of this line by looking at the relationship between the x and y coordinates of each point?
- Let us plot the points (− 3, 6), (− 2, 4), (0, 0), (1, − 2), (2, − 4), (3, − 6) in the coordinate plane on a graph paper as shown in the figure. Join the points (− 3, 6) and (3, − 6) using a ruler. Doing so, observe that all five points lie on a straight line. Can you guess the equation of this line by looking at the relationship between the x and y coordinates of each point?
- Draw the graphs of y = 12x, y = x, y = 2x by selecting suitable points on these lines.
(Hint: In order to graph y = 12x, we could take the points (0, 0) and (4, 2). Can you verify that these lie on the line?) - The figure shows all the three graphs on the same axes. Does this help you to conclude anything about the linear equation y = ax, a > 0 as a varies? What happens when a > 1 and when a < 1?
(Hint: You may also plot the equations y = 3x and y = 13x on the same axes.) - Now let us draw the graphs of y = −13x, y = −x, y = −3x by selecting suitable points on these lines.