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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17 questions

SA-9709-P1-CG2-0012 marksL1 FoundationCalculatorGradients
Find the equation of the line parallel to y=4x−1y=4x-1 that passes through (2,3)(2,3), in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0041 markL1 FoundationCalculatorGradients
Are the lines y=2x+1y=2x+1 and x+2y=5x+2y=5 perpendicular?
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-G011 markL1 FoundationCalculatorGradients
A line has gradient 25\tfrac25. State the gradient of a line perpendicular to it.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-Q011 markL1 FoundationCalculatorGradients
Two lines are perpendicular. The product of their gradients is
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-Q021 markL1 FoundationCalculatorGradients
Which line is parallel to y=3x+2y=3x+2?
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG-Q011 markL1 FoundationCalculatorGradients
A line is perpendicular to a line with gradient 23\tfrac23. Its gradient is
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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG-0024 marksL2 StandardCalculatorGradients
AA is (−2, 3)(-2,\,3) and BB is (4, 7)(4,\,7). Find the equation of the perpendicular bisector of ABAB, in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG-0133 marksL2 StandardCalculatorGradients
Find the equation of the line through (5, 1)(5,\,1) that is parallel to 2x−5y=32x-5y=3, in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0022 marksL2 StandardCalculatorGradients
Find the equation of the line through (1,4)(1,4) perpendicular to y=−13x+2y=-\tfrac13x+2, in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0033 marksL2 StandardCalculatorGradients
Find the equation of the line through (4,1)(4,1) perpendicular to 2x+3y=62x+3y=6, in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0052 marksL2 StandardCalculatorGradients
The lines y=kx+1y=kx+1 and y=14xy=\tfrac14x are perpendicular. Find kk.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0064 marksL2 StandardCalculatorBisectors
Find the equation of the perpendicular bisector of A(2,1)A(2,1) and B(6,9)B(6,9), in the form y=mx+cy=mx+c.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0082 marksL2 StandardCalculatorShapes
A(1,1)A(1,1), B(4,5)B(4,5) and C(8,2)C(8,2) form a triangle. At which vertex is there a right angle?
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0074 marksL3 ExamCalculatorBisectors
Find the coordinates of the foot of the perpendicular from P(5,5)P(5,5) to the line y=x−2y=x-2.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0093 marksL3 ExamCalculatorShapes
ABCDABCD is a parallelogram with A(0,0)A(0,0), B(4,2)B(4,2) and C(2,6)C(2,6). Find the coordinates of DD.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG-0155 marksL4 ChallengingCalculatorGradients
AA is (1, 1)(1,\,1) and BB is (7, 3)(7,\,3). The line through AA perpendicular to ABAB meets the yy-axis at CC. Find the yy-coordinate of CC and the area of triangle ABCABC.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P1-CG2-0106 marksL5 ExtensionCalculatorBisectors
AA is (1,2)(1,2) and BB is (5,6)(5,6). The perpendicular bisector of ABAB meets the xx-axis at PP. (i) Find the xx-coordinate of PP. (ii) Find the area of triangle ABPABP.

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Salam Academy Original Practice · Proposed mark scheme