Concepts and applications · 7 questions · 17 practice marks · 40 minutes. MCQs are scored automatically; written answers use step-wise self/teacher review.
Section A
Question 1(1 Mark)
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Full question paper and step-wise model answers
Practice marks are not official examination weightage. Written answers require self- or teacher assessment.
Question 1 · 1 mark
The point (−3,4) lies in:
Quadrant I
Quadrant II
Quadrant III
Quadrant IV
Model answer and reasoning
(b) Quadrant II
Its x-coordinate is negative and y-coordinate positive.
This sign combination identifies Quadrant II.
Question 2 · 1 mark
The distance between (1,2) and (4,6) is:
3
4
5
7
Model answer and reasoning
(c) 5
The horizontal and vertical changes are 3 and 4.
Distance = √(3² + 4²) = 5.
Question 3 · 3 marks
Explain how assigning coordinates to a geometric point connects geometry with algebra. On a grid, mark A(2,1) and B(−1,3), and describe B relative to A.
Model answer and reasoning
B is 3 units left and 2 units above A; coordinates allow geometric properties to be expressed algebraically.
Two perpendicular axes and a common origin fix the reference system.
An ordered pair records signed horizontal and vertical displacements; equations can then describe sets of geometric points.
B − A = (−1 − 2, 3 − 1) = (−3,2). Plot using these displacements.
Question 4 · 3 marks
Find the midpoint of A(−2,5) and B(6,−1). Verify it by comparing the displacements from each endpoint.
Model answer and reasoning
M = (2,2).
Average x-coordinates: (−2 + 6)/2 = 2.
Average y-coordinates: (5 − 1)/2 = 2.
M − A = (4,−3) and B − M = (4,−3), so M is equally placed along the segment.
Question 5 · 3 marks
A rectangular classroom has grid vertices (0,0), (8,0), (8,6), (0,6), with 1 grid unit = 1 metre. Mark the centre and find the straight-line distance from (0,0) to the opposite corner.
Model answer and reasoning
Centre (4,3); diagonal 10 m.
Draw a rectangle 8 units wide and 6 units high using the given coordinates.
The centre is the midpoint of opposite corners: ((0 + 8)/2, (0 + 6)/2) = (4,3).
The diagonal length is √(8² + 6²) = √100 = 10 m.
Question 6 · 3 marks
Use distances to decide whether A(0,0), B(2,2) and C(5,5) are collinear.
Model answer and reasoning
They are collinear, with B between A and C.
AB = √8 = 2√2; BC = √18 = 3√2.
AC = √50 = 5√2.
AB + BC = AC, so equality in the triangle inequality places B on segment AC.
Question 7 · 3 marks
Check whether A(0,0), B(4,0), C(0,3) form a right-angled triangle. Identify the right-angle vertex.
Model answer and reasoning
The triangle is right-angled at A.
AB² = 16 and AC² = 9.
BC² = (4 − 0)² + (0 − 3)² = 25.
AB² + AC² = BC²; by the converse of Pythagoras the angle opposite BC, at A, is 90°.