Question 1: Given a point (3, 4) and a gradient m = 2, write the equation of the line using y - y₁ = m(x - x₁).

Answer: y - 4 = 2(x - 3)

Question 2: Plot the line with equation y - 1 = 3(x - 2) on the grid.

Answer: Student draws line passing through (2,1) with gradient 3

Question 3: Identify the point (x, y) on the line y - 5 = -1/2(x - 4) when x = 6.

Answer: y - 5 = -1/2(6 - 4) y - 5 = -1/2 * 2 y - 5 = -1 y = 4

Question 1: A line passes through points (2, 3) and (4, 7). Write the equation of the line in y - y₁ = m(x - x₁) form, using (2,3) as the point.

Answer: y - 3 = 2(x - 2)

Question 2: A student writes the equation y + 2 = 1/3(x - 4). Is this correct? If not, identify and correct the mistake.

Answer: Incorrect because the sign should be negative: y - (-2) = 1/3(x - 4) or y + 2 = 1/3(x - 4) is correct if the point is (4, -2). The mistake is in misapplying the sign.

Question 3: Explain why using y - y₁ = m(x - x₁) is important when constructing the equation of a line given a point and gradient.

Answer: It ensures the line passes through the given point and has the correct gradient, maintaining the relationship between x and y coordinates.

Question 1: A cyclist starts at point (1, 2) and rides with a constant rate of 4 units for every 1 unit increase in x. Write the equation of the cyclist's path and plot it on the grid.

Answer: y - 2 = 4(x - 1)

Question 1: Given the line passes through (5, -1) and the equation in y - y₁ = m(x - x₁) form is y + 1 = -2(x - 5). Find the point where this line crosses the y-axis.

Answer: Set x=0: y + 1 = -2(0 - 5) y + 1 = -2 * (-5) y + 1 = 10 y = 9

Question 2: Construct a line on the grid that passes through (3, 2) with gradient -1/2. Write the equation and explain your steps.

Answer: Using point (3,2): y - 2 = -1/2(x - 3). This ensures the line passes through the point and has the specified gradient.

Question 1: A student writes the equation y + 3 = 2(x - 1) after being given point (1, -3) and gradient m=2. Is this correct? If not, fix the error.

Answer: Incorrect sign; it should be y - (-3) = 2(x - 1) or y + 3 = 2(x - 1). The student used a plus sign instead of minus for y - y₁.

Question 2: Identify the mistake in the following statement: 'To find the equation, I just substitute the point and gradient into y - y₁ = m(x - x₁)', when the point is (4, 5) and m=0.5.

Answer: No mistake; the process is correct. The student should write y - 5 = 0.5(x - 4).