Secondary 1 Mathematics Graphs Coordinate Geometry Quiz

Secondary 1 Mathematics Quiz - Graphs Coordinate Geometry — Answer Key

Total Marks: 40

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Section A: Coordinate Basics and Plotting

1. [2 marks]

Point A: (−3, 2)
Point B: (0, 4)
Point C: (2, −3)
Point D: (0, 0)

Marking: ½ mark each

Teaching note: The first number in an ordered pair (x, y) is the x-coordinate (horizontal position; how far left or right from the origin). The second number is the y-coordinate (vertical position; how far up or down). For point A: move 3 units left (negative) and 2 units up (positive) → quadrant II. Point D is at the origin where both coordinates are zero. Common mistake: writing coordinates as (y, x) instead of (x, y).

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2. [2 marks]

Correct plots with labels:

• P(4, 2): 4 right, 2 up
• Q(−2, 5): 2 left, 5 up
• R(0, −3): on y-axis, 3 down
• S(−4, −1): 4 left, 1 down

Marking: ½ mark each point correctly plotted and labeled

Teaching note: To plot a point (a, b): start at origin, move a units horizontally (right if positive, left if negative), then move b units vertically (up if positive, down if negative). The point R with x = 0 lies on the y-axis, not in any quadrant.

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3. [2 marks]

(a) IV (or fourth quadrant) — x positive, y negative
(b) x-axis — y = 0 means on the x-axis, not in a quadrant
(c) y-axis — x = 0 means on the y-axis
(d) III (or third quadrant) — both x and y negative

Marking: ½ mark each

Teaching note: Quadrant numbering goes counter-clockwise from the positive x-axis: I (+,+), II (−,+), III (−,−), IV (+,−). A point on an axis is not in any quadrant. Common trap: saying (0, 6) is in quadrant I or on the x-axis.

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4. [2 marks]

Since T(3, k) lies on y = 2x − 5, substitute x = 3:

k = 2(3) − 5
k = 6 − 5
k = 1

Marking: 1 mark method (substitution), 1 mark correct answer

Teaching note: "Lies on the line" means the point's coordinates satisfy the equation. Replace x with 3 and y with k, then solve. Always substitute carefully: 2 × 3 = 6, not 23.

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5. [2 marks]

Since U(m, 4) lies on y = −x + 7, substitute y = 4:

4 = −m + 7
m = 7 − 4
m = 3

Marking: 1 mark method (substitution and rearrangement), 1 mark correct answer

Teaching note: Here we know the y-value and need to find x. Substitute y = 4 first, then solve for m. Watch negatives: −m + 7 = 4 means m = 3 (not −3). Check: (−3) + 7 = 4 ✓.

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Section B: Straight Line Graphs and Gradients

6. [2 marks]

Gradient = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 5}{6 - 2} = \frac{8}{4} =$ 2

Marking: 1 mark formula/method, 1 mark answer

Teaching note: The gradient (or slope) measures steepness: change in y divided by change in x. Always subtract in the same order: (y-value of second point minus y-value of first point) over (matching x-values). The gradient 2 means "for every 1 unit right, go 2 units up."

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7. [2 marks]

Gradient = $\frac{-4 - 4}{1 - (-3)} = \frac{-8}{4} =$ −2

Marking: 1 mark method (correct handling of negatives), 1 mark answer

Teaching note: Be careful with double negatives: 1 − (−3) = 1 + 3 = 4. The negative gradient means the line slopes downhill from left to right. Common error: calculating $\frac{4 - (-4)}{-3 - 1} = \frac{8}{-4} = -2$ is also correct if consistent, but mixing orders gives wrong signs.

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8. [2 marks]

The equation is y = −3x + 5

Marking: 1 mark for m = −3, 1 mark for c = 5 with correct format

Teaching note: The form y = mx + c is called slope-intercept form. m is the gradient, c is the y-intercept (where the line crosses the y-axis, at point (0, c)). When x = 0, y = c, so (0, 5) directly gives c = 5.

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9. [2 marks]

Substitute x = 3, y = 8 into y = 2x + c:

8 = 2(3) + c
8 = 6 + c
c = 2

Equation: y = 2x + 2

Marking: 1 mark finding c, 1 mark writing full equation

Teaching note: "Passes through" means the point satisfies the equation. We know m = 2 and need c. Substitute the known point to find the unknown c. Always write the final equation with both m and c substituted in.

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10. [2 marks]

For x-intercept: set y = 0: 0 = 3x − 6, so x = 2. x-intercept = 2 (or point (2, 0))

For y-intercept: set x = 0: y = 3(0) − 6 = −6. y-intercept = −6 (or point (0, −6))

Marking: 1 mark each; must be correct values or correct coordinate forms

Teaching note: x-intercept: where line crosses x-axis (y = 0). y-intercept: where line crosses y-axis (x = 0). For y = mx + c, the y-intercept is always c. Here c = −6, confirming our answer.

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11. [2 marks]

(a) Gradient = $\frac{6 - (-2)}{4 - 0} = \frac{8}{4} =$ 2

(b) y = 2x − 2 (since c = −2 from point (0, −2))

Marking: (a) 1 mark, (b) 1 mark

Teaching note: The point (0, −2) is on the y-axis, so it's the y-intercept, giving c = −2 directly. Alternatively, use y — y₁ = m(x — x₁) with either point: y — 6 = 2(x — 4) → y = 2x — 2.

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12. [2 marks]

Substitute x = 5, y = 11 into y = 2x + 1:

Left side = 11
Right side = 2(5) + 1 = 10 + 1 = 11

Since LHS = RHS, yes, the point (5, 11) lies on the line.

Marking: 1 mark substitution, 1 mark conclusion with clear equality check

Teaching note: Never just say "yes" — show the check. Substitute both coordinates and verify both sides equal. Common mistake: only substituting x and forgetting to compare with the given y-value.

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Section C: Drawing and Analyzing Graphs

13. [2 marks]

x	−2	0	2	4
y	−7	−3	1	5

Calculated: y = 2(−2) − 3 = −7; y = 2(0) − 3 = −3; y = 2(2) − 3 = 1; y = 2(4) − 3 = 5

Straight line drawn through (−2, −7), (0, −3), (2, 1), (4, 5) with ruler.

Marking: 1 mark table, 1 mark correct line

Teaching note: Always calculate at least 3 points to check, though 2 determine a line. The third point verifies. Use a ruler, and extend the line slightly beyond your plotted points. Common error: arithmetic with negatives, e.g., 2(−2) = −4, then −4 − 3 = −7.

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14. [2 marks]

(a) Gradient = $\frac{60 - 0}{4 - 0} = \frac{60}{4} =$ 15

(b) The gradient represents the speed of the cyclist in km/h (or "the cyclist travels 15 km per hour").

Marking: (a) 1 mark, (b) 1 mark for correct interpretation with unit

Teaching note: In a distance-time graph, gradient = speed. The steeper the line, the faster. Here, in 4 hours the cyclist travels 60 km, so 15 km each hour. Common error: saying "distance" or "time" without relating them as a rate.

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15. [2 marks]

(a) From (0, 4): b = 4 (y-intercept)

Gradient a = $\frac{10 - 4}{2 - 0} = \frac{6}{2} =$ 3

So a = 3, b = 4; equation is y = 3x + 4

(b) For x-intercept: set y = 0: 0 = 3x + 4, so x = −4/3

Marking: (a) ¾ mark (½ for b, ¼ for a), (b) ¼ mark; or equivalent breakdown

Teaching note: a is the gradient, b is the y-intercept. Read b directly from where the line cuts the y-axis. For the x-intercept, always set y = 0 and solve. The negative x-intercept means the line crosses left of origin.

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16. [2 marks]

Both lines should pass through (0, 2) — the common y-intercept.

Line (a) y = x + 2: gradient 1, passes through (0, 2), (−2, 0), (2, 4) etc. — slopes up to right

Line (b) y = −x + 2: gradient −1, passes through (0, 2), (−2, 4), (2, 0) etc. — slopes down to right

Both labeled correctly on grid.

Marking: 1 mark each line (correct gradient direction and y-intercept)

Teaching note: Lines with same c, different m intersect on the y-axis at (0, c). The positive gradient line goes up; negative goes down. They form a "V" shape. Check: at x = 0, both give y = 2. At x = 2: (a) gives 4, (b) gives 0.

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17. [2 marks]

(a) C = 3 + 2(8) = 3 + 16 = $19

(b) The 3 represents the flag fall (or basic hire charge / initial fixed cost) — the cost before any distance is travelled, in dollars.

Marking: (a) 1 mark, (b) 1 mark with clear explanation

Teaching note: In y = mx + c applied to real contexts: c is the fixed amount, m is the rate per unit. Here, $3 pays for getting in the taxi, then$2 per km. When n = 0, C = 3. Always state what the number represents in context, not just "y-intercept."

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Section D: Problem Solving and Application

18. [2 marks]

(a) Rectangle (or "square" — but check: length 4, width 3, so rectangle)

(b) Length = 5 − 1 = 4 units, Width = 5 − 2 = 3 units

Area = 4 × 3 = 12 square units

Marking: (a) 1 mark, (b) 1 mark

Teaching note: Plot mentally or sketch: A(1,2) to B(5,2) is horizontal, length 4. B(5,2) to C(5,5) is vertical, length 3. Opposite sides equal, all angles 90° → rectangle. Area = length × width. Common error: thinking it's a square (would need sides equal).

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19. [2 marks]

(a) For y-intercept: set x = 0: 2(0) + 3y = 12, so 3y = 12, y = 4. Coordinates: (0, 4)

(b) Rearrange to y = mx + c form:

2x + 3y = 12
3y = −2x + 12
y = −$\frac{2}{3}$x + 4

Gradient = −2/3

Marking: (a) 1 mark, (b) 1 mark

Teaching note: To find gradient from general form (ax + by = c): rearrange to make y the subject. The coefficient of x becomes the gradient. Or use formula: gradient = −a/b = −2/3. Common error: saying gradient is 2 or 2/3 (forgetting the negative or dividing wrong way).

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20. [2 marks]

(a) P(1, 1)

(b) Gradient of L₁ = $\frac{3 - (-1)}{2 - 0} = \frac{4}{2} = 2$; c = −1 from (0, −1)

Equation: y = 2x − 1

(c) Gradient of L₂ = $\frac{2 - 5}{3 - 0} = \frac{-3}{3} = −1$; c = 5 from (0, 5)

Equation: y = −x + 5

Marking: (a) ½ mark, (b) 1 mark, (c) ½ mark

Teaching note: For intersection point: read coordinates directly from graph where lines cross — check x then y. For equations: find gradient using two given points on each line, then use c from y-intercept. Verify: does P(1,1) satisfy both? L₁: 1 = 2(1) − 1 = 1 ✓; L₂: 1 = −1 + 5 = 4? No — wait, let me recheck: if L₂ passes through (3,2) and (0,5): yes, 2 = −3 + 5 = 2 ✓. At P(1,1): 1 = −1 + 5 = 4? This seems inconsistent.

Actually with the given points (0,5) and (3,2): equation is y = −x + 5. At x = 1: y = 4, not 1. The intersection P(1,1) with L₁: y = 2(1) − 1 = 1 ✓. But L₂ at (1,1): 1 ≠ −1+5 = 4.

Correction for teaching: The image shows P at intersection; if L₂ is y = −x + 5, then P should be where 2x − 1 = −x + 5: 3x = 6, x = 2, P(2,3). Given the image data has P(1,1), there may be a slight inconsistency in the problem numbers. In a real exam, the diagram would be drawn precisely. Students should read P from the diagram for part (a), then derive equations from their respective given points for (b) and (c).

For answer key purposes with given data: (a) P(1, 1) from diagram. For equations, use the two given points on each line as stated. The slight numerical inconsistency is a teaching moment: diagrams may be approximate; use given coordinates for equations, read intersection visually.

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End of Answer Key