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3 marksStraight line equation4MA1 style ✏️ No calculator
Find the equation of the line passing through (2, 5) and (6, 13). Give your answer in the form y = mx + c. (3 marks)
Hint: Step 1: gradient = (y₂−y₁)/(x₂−x₁). Step 2: substitute gradient and one point into y−y₁ = m(x−x₁). Step 3: rearrange to y = mx+c.
+30 XP
3 marksPerpendicular linesGrade 7 Booklet style ✏️ No calculator
Line L has equation y = 3x − 2. Line M is perpendicular to L and passes through (6, 1). Find the equation of line M. (3 marks)
Hint: Perpendicular gradient = −1/m. Gradient of L is 3, so gradient of M is −1/3. Then use y−y₁ = m(x−x₁) with the given point.
+30 XP
3 marksGraph transformationGrade 7 Booklet style ✏️ No calculator
The graph of y = f(x) passes through (3, 4). (a) What are the coordinates of this point on y = f(x+2)? (b) On y = 2f(x)? (c) On y = f(−x)? (3 marks)
Hint: (a) f(x+2) shifts left by 2 — subtract 2 from x-coordinate. (b) 2f(x) doubles the y-coordinate. (c) f(−x) reflects in y-axis — negate x-coordinate.
+30 XP
1 markGradient of perpendicular4MA1 style
A line has gradient 2/5. What is the gradient of a line perpendicular to it?