Salam Academy
Practise

Question bank

Original exam-style questions, mapped to your syllabus. Answers are checked for mathematical equivalence and every attempt updates your mastery.

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12 questions

SA-4024-CG-0042 marksL1 FoundationNon-calculatorParallel
Find the equation of the line parallel to y=2x−7y=2x-7 that passes through (3,1)(3,1).

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0012 marksL1 FoundationCalculatorParallel
The line parallel to y=3x+1y=3x+1 through (2,4)(2,4) has equation y=3x+cy=3x+c. Find cc.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0021 markL1 FoundationCalculatorPerpendicular
A line has gradient −4-4. Write down the gradient of a line perpendicular to it.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0041 markL1 FoundationCalculatorParallel
Which line is parallel to 2y=4x+32y=4x+3?
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG-Q011 markL1 FoundationCalculatorPerpendicular
A line has gradient 23\tfrac23. A line perpendicular to it has gradient
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG-0053 marksL2 StandardNon-calculatorPerpendicular
Find the equation of the line perpendicular to y=2x+1y=2x+1 that passes through (4,3)(4,3).

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0033 marksL2 StandardCalculatorPerpendicular
The line perpendicular to y=−2x+5y=-2x+5 through (4,1)(4,1) has equation y=mx+cy=mx+c. Find mm and cc.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0051 markL2 StandardCalculatorPerpendicular
Are the lines y=3x+2y=3x+2 and x+3y=6x+3y=6 perpendicular?
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0062 marksL2 StandardCalculatorParallel
The line y=kx+1y=kx+1 is parallel to the line 4x−2y=54x-2y=5. Find kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG-0064 marksL3 ExamNon-calculatorPerpendicular
Find the equation of the perpendicular bisector of A(−2,1)A(-2,1) and B(4,5)B(4,5), in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG2-0074 marksL3 ExamCalculatorPerpendicular
AA is (0,0)(0,0) and BB is (4,2)(4,2). The perpendicular bisector of ABAB has equation y=−2x+cy=-2x+c. Find cc.

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Salam Academy Original Practice · Proposed mark scheme
SA-4024-CG-0166 marksL4 ChallengingNon-calculatorPerpendicular
AA is (1,2)(1,2), BB is (7,4)(7,4) and CC is (3,8)(3,8). (i) Find the equation of the line through CC perpendicular to ABAB. (ii) Find the coordinates of the point where this line meets ABAB.

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Salam Academy Original Practice · Proposed mark scheme