Secondary 3 Elementary Mathematics Graphs Coordinate Geometry Quiz

Secondary 3 Elementary Mathematics Quiz — Answer Key

Topic: Graphs Coordinate Geometry
Total Marks: 40

Section A: Coordinate Geometry — Distance, Gradient, and Midpoint

Question 1 [6 marks]

(a) Gradient of AB = (13 − 5) / (8 − 2) = 8 / 6 = 4/3
[2 marks] — 1 mark for correct substitution into gradient formula, 1 mark for correct simplified answer.

(b) Length of AB = √[(8 − 2)² + (13 − 5)²] = √[36 + 64] = √100 = 10 units
[2 marks] — 1 mark for correct substitution, 1 mark for correct answer.
Note: The question asks for simplified surd form; since √100 = 10, the answer 10 is acceptable. If the numbers had produced a non-perfect square, students should leave it as a surd.

(c) Midpoint of AB = ((2 + 8)/2, (5 + 13)/2) = (5, 9)
[2 marks] — 1 mark for correct x-coordinate, 1 mark for correct y-coordinate.

Question 2 [3 marks]

Gradient = (7 − k) / (5 − (−3)) = (7 − k) / 8 = ¾

7 − k = 6
k = 1

[3 marks] — 1 mark for setting up the gradient equation, 1 mark for solving correctly, 1 mark for final answer.

Common mistake: Students may subtract in the wrong order (k − 7 instead of 7 − k), giving k = 13. Award method marks if the setup is otherwise correct.

Question 3 [3 marks]

Since B is the midpoint of AC:
Midpoint formula: ((1 + 4)/2, (2 + m)/2) = (7, 4)

x-coordinate: (1 + 4)/2 = 5/2 ≠ 7 — this does not match.
Re-reading the problem: B(7, 4) is the midpoint of A(1, 2) and C(4, m).

x-coordinate: (1 + 4)/2 = 5/2. This should equal 7.
(1 + 4)/2 = 2.5 ≠ 7. This is inconsistent.

Correction: Let us re-examine. If B(7, 4) is the midpoint of A(1, 2) and C(4, m):
x: (1 + 4)/2 = 2.5, but B has x = 7. This is a contradiction.

Revised interpretation: The problem should be read as: B is the midpoint of AC, so:
(1 + 4)/2 = 7 → 5/2 = 7, which is impossible.

Let me re-read: A(1, 2), B(7, 4), C(4, m). B is the midpoint of AC.
(1 + 4)/2 = 2.5 ≠ 7. This is inconsistent.

Alternative reading: Perhaps C has coordinates (c, m) and we need to find both. But the question says C(4, m). Let me check if the problem might mean A is the midpoint, or if the coordinates are different.

Re-examining the question as written: Given the coordinates as stated, there is an inconsistency. However, for the purpose of this quiz, let us proceed with the y-coordinate calculation, which is solvable:

y-coordinate: (2 + m)/2 = 4
2 + m = 8
m = 4

Wait — (2 + m)/2 = 4 gives 2 + m = 8, so m = 6.

[3 marks] — 1 mark for using the midpoint formula for the y-coordinate, 1 mark for correct equation, 1 mark for m = 6.

Note: There appears to be an inconsistency in the x-coordinates as given. In an exam setting, the question would be checked for internal consistency. For this quiz, students should proceed with the y-coordinate calculation: (2 + m)/2 = 4, giving m = 6.

Question 4 [3 marks]

Since D(−2, 1), E(4, 7), and F(p, 3) are collinear, the gradient of DE equals the gradient of EF (or DF).

Gradient of DE = (7 − 1) / (4 − (−2)) = 6 / 6 = 1

Gradient of DF = (3 − 1) / (p − (−2)) = 2 / (p + 2)

Set equal to 1:
2 / (p + 2) = 1
p + 2 = 2
p = 0

[3 marks] — 1 mark for finding gradient of DE, 1 mark for setting up equation, 1 mark for p = 0.

Question 5 [5 marks]

(a) Using y = mx + c with m = −2 and point (6, −1):
−1 = −2(6) + c
−1 = −12 + c
c = 11

Equation: y = −2x + 11
[2 marks] — 1 mark for substituting into y = mx + c, 1 mark for correct equation.

(b) Substitute x = 3 into the equation:
y = −2(3) + 11 = −6 + 11 = 5

Since y = 5 when x = 3, the point (3, 5) does lie on the line.
[3 marks] — 1 mark for substituting x = 3, 1 mark for finding y = 5, 1 mark for correct conclusion with reasoning.

Section B: Equation of a Straight Line and Parallel/Perpendicular Lines

Question 6 [4 marks]

Gradient = (−2 − 4) / (3 − (−1)) = −6 / 4 = −3/2

Using point (−1, 4):
y − 4 = −3/2(x + 1)
y − 4 = −3/2 x − 3/2
Multiply through by 2:
2y − 8 = −3x − 3
3x + 2y − 5 = 0

[4 marks] — 1 mark for correct gradient, 1 mark for correct substitution into point-gradient form, 1 mark for correct rearrangement, 1 mark for final answer in required form with integer coefficients.

Question 7 [5 marks]

(a) Rearrange 3x − 4y + 8 = 0:
4y = 3x + 8
y = 3/4 x + 2

Gradient of L₁ = 3/4
[2 marks] — 1 mark for rearranging, 1 mark for correct gradient.

(b) L₂ is parallel to L₁, so gradient of L₂ = 3/4.
L₂ passes through (2, −5):
y − (−5) = 3/4(x − 2)
y + 5 = 3/4 x − 3/2
y = 3/4 x − 3/2 − 5
y = 3/4 x − 13/2

Equation: y = 3/4 x − 13/2
[3 marks] — 1 mark for using correct gradient, 1 mark for correct substitution, 1 mark for correct simplified equation.

Question 8 [6 marks]

(a) Gradient of L₃ = −½.
Since L₄ ⊥ L₃: gradient of L₄ = 2 (negative reciprocal of −½).
[1 mark]

(b) L₄ passes through (4, 1) with gradient 2:
y − 1 = 2(x − 4)
y − 1 = 2x − 8
y = 2x − 7
[2 marks] — 1 mark for substitution, 1 mark for correct equation.

(c) At the point of intersection, the y-values are equal:
−½x + 3 = 2x − 7
3 + 7 = 2x + ½x
10 = 5/2 x
x = 4

Substitute x = 4 into y = 2x − 7:
y = 8 − 7 = 1

Point of intersection = (4, 1)
[3 marks] — 1 mark for equating the two expressions, 1 mark for solving x = 4, 1 mark for finding y and stating coordinates.

Note: The point of intersection is (4, 1), which is the given point on L₄. This is consistent since (4, 1) also lies on L₃: y = −½(4) + 3 = −2 + 3 = 1. ✓

Question 9 [5 marks]

(a) For point A (x-intercept), set y = 0:
2x + 5(0) = 10
2x = 10
x = 5
A = (5, 0)

For point B (y-intercept), set x = 0:
2(0) + 5y = 10
5y = 10
y = 2
B = (0, 2)

[2 marks] — 1 mark for each correct point.

(b) Triangle OAB has base OA = 5 and height OB = 2.
Area = ½ × 5 × 2 = 5 square units
[3 marks] — 1 mark for identifying base and height, 1 mark for correct formula, 1 mark for correct answer.

Question 10 [3 marks]

Since P(0, 3), Q(4, 7), and R(2, t) are collinear, gradient of PQ = gradient of PR.

Gradient of PQ = (7 − 3) / (4 − 0) = 4/4 = 1

Gradient of PR = (t − 3) / (2 − 0) = (t − 3) / 2

Set equal to 1:
(t − 3) / 2 = 1
t − 3 = 2
t = 5

[3 marks] — 1 mark for gradient of PQ, 1 mark for setting up equation, 1 mark for t = 5.

Section C: Graphs of Functions and Gradient of a Curve

Question 11 [6 marks]

(a) When x = 0: y = 0 − 0 + 3 = 3
Crosses y-axis at (0, 3)
[1 mark]

(b) Set y = 0: x² − 4x + 3 = 0
(x − 1)(x − 3) = 0
x = 1 or x = 3

Crosses x-axis at (1, 0) and (3, 0)
[2 marks] — 1 mark for setting up equation, 1 mark for both correct points.

(c) Line of symmetry: x = −(−4)/(2×1) = 4/2 = 2
x = 2
[1 mark]

(d) Minimum point occurs at x = 2:
y = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1
Minimum point = (2, −1)
[2 marks] — 1 mark for x = 2, 1 mark for y = −1 and correct coordinates.

Question 12 [4 marks]

(a) Set y = 0: (x − 2)(x + 4) = 0
x = 2 or x = −4

Crosses x-axis at (−4, 0) and (2, 0)
[2 marks] — 1 mark for each correct point.

(b) The line of symmetry is midway between the x-intercepts:
x = (−4 + 2) / 2 = −1

When x = −1: y = (−1 − 2)(−1 + 4) = (−3)(3) = −9

Minimum point = (−1, −9)
[2 marks] — 1 mark for x = −1, 1 mark for y = −9 and correct coordinates.

Question 13 [5 marks]

(a) Completed table:

[2 marks] — 1 mark for each correct pair (deduct ½ per error, minimum 0).

(b) Students should draw a smooth hyperbola with two branches, one in the first quadrant and one in the third quadrant, with asymptotes along the axes.
[2 marks] — 1 mark for correct general shape, 1 mark for correct positioning of branches.

(c) Lines of symmetry: y = x and y = −x
[1 mark] — Both required for the mark.

Question 14 [5 marks]

(a) Completed table:

[2 marks] — ½ mark for each correct value (deduct per error, minimum 0).

(b) Asymptote: y = 0 (the x-axis)
[1 mark]

(c) As x becomes very large, y increases exponentially (y → ∞). The graph becomes steeper and steeper.
[2 marks] — 1 mark for stating that y increases/grows, 1 mark for describing exponential growth or that y becomes very large.

Question 15 [4 marks]

(a) Students should draw a tangent to the curve at (3, 0). A reasonable estimate of the gradient is obtained by calculating the gradient of the tangent line drawn.

From the equation y = x² − 2x − 3, the derivative dy/dx = 2x − 2.
At x = 3: gradient = 2(3) − 2 = 4

Students' estimates from their drawn tangent should be close to 4 (accept 3.5 to 4.5).
[2 marks] — 1 mark for correct tangent construction, 1 mark for reasonable estimate near 4.

(b) When x = 2: y = 4 − 4 − 3 = −3, so point is (2, −3).
When x = 4: y = 16 − 8 − 3 = 5, so point is (4, 5).

Gradient of chord = (5 − (−3)) / (4 − 2) = 8 / 2 = 4
[2 marks] — 1 mark for correct y-values, 1 mark for correct gradient.

Section D: Applications and Problem Solving

Question 16 [7 marks]

(a) Gradient of AB = (−4 − 8) / (5 − 1) = −12 / 4 = −3

Using point A(1, 8):
y − 8 = −3(x − 1)
y − 8 = −3x + 3
y = −3x + 11
[2 marks] — 1 mark for correct gradient, 1 mark for correct equation.

(b) At the x-intercept, y = 0:
0 = −3x + 11
3x = 11
x = 11/3

C = (11/3, 0)
[2 marks] — 1 mark for setting y = 0, 1 mark for correct x-coordinate.

(c) The perpendicular line through C has gradient = 1/3 (negative reciprocal of −3).

Using point C(11/3, 0):
y − 0 = 1/3(x − 11/3)
y = 1/3 x − 11/9

Multiply through by 9: 9y = 3x − 11
3x − 9y − 11 = 0 (or y = 1/3 x − 11/9)
[3 marks] — 1 mark for correct perpendicular gradient, 1 mark for substitution, 1 mark for correct equation.

Question 17 [5 marks]

(a) PQ is horizontal from (−3, 1) to (5, 1), so length of PQ = 5 − (−3) = 8.
The perpendicular height from R(2, 6) to line PQ (which is y = 1) is 6 − 1 = 5.

Area = ½ × 8 × 5 = 20 square units
[3 marks] — 1 mark for base length, 1 mark for height, 1 mark for correct area.

(b) The length of the perpendicular from R to PQ is simply the vertical distance:
|6 − 1| = 5 units
[2 marks] — 1 mark for method, 1 mark for correct answer.

Question 18 [6 marks]

(a) When t = 0: d = −60(0) + 180 = 180 km
[1 mark]

(b) At the checkpoint, d = 0:
0 = −60t + 180
60t = 180
t = 3 hours
[2 marks] — 1 mark for setting d = 0, 1 mark for t = 3.

(c) The gradient is −60. This represents the speed of the car, which is 60 km/h travelling towards the checkpoint (the negative sign indicates the distance is decreasing).
[2 marks] — 1 mark for identifying the gradient as −60, 1 mark for explaining it represents speed/direction.

(d) When t = 2.5: d = −60(2.5) + 180 = −150 + 180 = 30 km
[1 mark]

Question 19 [6 marks]

(a) The parabola passes through (0, −5) and (4, −5). Since both points have the same y-value, the line of symmetry is midway between them:
x = (0 + 4) / 2 = 2
[2 marks] — 1 mark for reasoning, 1 mark for x = 2.

(b) From (a), the line of symmetry is x = 2, so −b/(2×1) = 2, giving b = −4.

The curve passes through (0, −5):
−5 = 0 + 0 + c
c = −5

b = −4, c = −5
[2 marks] — 1 mark for b, 1 mark for c.

(c) The minimum value occurs at x = 2:
y = (2)² + (−4)(2) + (−5) = 4 − 8 − 5 = −9
[2 marks] — 1 mark for substituting x = 2, 1 mark for correct minimum value.

Question 20 [7 marks]

(a) Midpoint of AC = ((−2 + 4)/2, (3 + (−1))/2) = (1, 1)
[2 marks] — 1 mark for x-coordinate, 1 mark for y-coordinate.

(b) In a parallelogram, diagonals bisect each other, so the midpoint of BD is also (1, 1).

Let D = (x, y). Midpoint of BD: ((6 + x)/2, (7 + y)/2) = (1, 1)

(6 + x)/2 = 1 → 6 + x = 2 → x = −4
(7 + y)/2 = 1 → 7 + y = 2 → y = −5

D = (−4, −5)
[3 marks] — 1 mark for using midpoint property, 1 mark for each correct coordinate.

(c) Diagonal BD passes through B(6, 7) and D(−4, −5).

Gradient = (−5 − 7) / (−4 − 6) = −12 / −10 = 6/5

Using point B(6, 7):
y − 7 = 6/5(x − 6)
y − 7 = 6/5 x − 36/5
y = 6/5 x − 36/5 + 35/5
y = 6/5 x − 1/5

Or in integer form: 5y = 6x − 1, i.e., 6x − 5y − 1 = 0
[2 marks] — 1 mark for correct gradient, 1 mark for correct equation.

End of Answer Key