Solutions — Graphing Linear Functions
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Q1 — Identify gradient and y-intercept
(a) y = 3x − 4: Already in y = mx + c form. m = 3, c = −4.
(b) y = −2x + 7: m = −2, c = 7.
(c) 2x + y = 5: Rearrange: y = −2x + 5. m = −2, c = 5.
(d) 3x − 4y + 8 = 0: Rearrange: 4y = 3x + 8 → y = (3/4)x + 2. m = 3/4 = 0.75, c = 2.
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Q2 — Gradient from two points
(a) (1, 3) and (4, 9): m = (9 − 3)/(4 − 1) = 6/3 = 2
(b) (−2, 5) and (3, −5): m = (−5 − 5)/(3 − (−2)) = −10/5 = −2
(c) (0, −3) and (6, 0): m = (0 − (−3))/(6 − 0) = 3/6 = 1/2 = 0.5
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Q3 — Write the equation of a line
(a) m = 4, c = −2: Direct substitution into y = mx + c: y = 4x − 2
(b) m = −½, through (4, 3):
Use point–gradient form: y − 3 = −½(x − 4)
y − 3 = −½x + 2
y = −½x + 5
Check: at (4, 3): y = −2 + 5 = 3 ✓
(c) Through (2, −1) and (5, 8):
Step 1 — gradient: m = (8 − (−1))/(5 − 2) = 9/3 = 3
Step 2 — use point (2, −1): y − (−1) = 3(x − 2)
y + 1 = 3x − 6
y = 3x − 7
Check: at (5, 8): y = 15 − 7 = 8 ✓
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Q4 — Features of y = −3x + 6
(a) Intercepts:
y-intercept: set x = 0: y = 6 → (0, 6)
x-intercept: set y = 0: 0 = −3x + 6 → 3x = 6 → x = 2 → (2, 0)
(b) Key features for sketch:
Plot the y-intercept at (0, 6) and the x-intercept at (2, 0). Draw a straight line through these two points and extend in both directions. The line has a negative slope, so it falls from left to right.
(c) Gradient = −3. This means for every 1 unit increase in x, y decreases by 3 units. As a rate of change, y decreases at a rate of 3 per unit of x.
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Q5 — Parallel, perpendicular or neither
(a) y = 2x + 3 and y = 2x − 5: Both have m = 2. Same gradient, different y-intercepts. Parallel.
(b) y = 4x − 1 and y = −¼x + 2: m&sub1; = 4, m&sub2; = −¼. Product: 4 × (−¼) = −1. Perpendicular.
(c) y = 3x + 1 and y = −3x + 1: m&sub1; = 3, m&sub2; = −3. Product: 3 × (−3) = −9 ≠ −1. Gradients are not equal. Neither. (Note: the lines are “reflections” of each other but not perpendicular.)
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Q6 — Parallel and perpendicular lines
(a) Parallel to y = 3x − 2 through (1, 4):
Parallel → same gradient m = 3.
y − 4 = 3(x − 1) → y = 3x + 1
y = 3x + 1
Check: (1, 4) gives y = 3 + 1 = 4 ✓
(b) Perpendicular to y = 2x + 1 through (6, 3):
Original m = 2. Perpendicular m⊥ = −1/2.
y − 3 = −½(x − 6) → y − 3 = −½x + 3 → y = −½x + 6
y = −½x + 6
Check: (6, 3) gives y = −3 + 6 = 3 ✓
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Q7 — Line through (−3, 7) and (1, −1)
(a) Gradient: m = (−1 − 7)/(1 − (−3)) = −8/4 = −2
(b) y-intercept: Use y − 7 = −2(x − (−3)):
y − 7 = −2(x + 3) = −2x − 6
y = −2x + 1, so y-intercept = 1 (point (0, 1))
(c) x-intercept: Set y = 0: 0 = −2x + 1 → 2x = 1 → x = ½
x-intercept: (0.5, 0)
(d) General form: From y = −2x + 1:
2x + y − 1 = 0
2x + y − 1 = 0
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Q8 — Graph interpretation: phone credit
(a) y-intercept = 8: This represents the initial credit before any calls are made — $8 of credit to start.
(b) Gradient = −2: The credit decreases by $2 for each call made. Each call costs $2.
(c) Credit runs out when y = 0: 0 = −2x + 8 → 2x = 8 → x = 4. Credit runs out after 4 calls.
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Q9 — Midpoint and perpendicular bisector
(a) Coordinates of B:
Midpoint formula: M = ((x_A + x_B)/2, (y_A + y_B)/2) = (3, 1)
(−1 + x_B)/2 = 3 → −1 + x_B = 6 → x_B = 7
(4 + y_B)/2 = 1 → 4 + y_B = 2 → y_B = −2
B = (7, −2)
(b) Gradient of AB:
m = (−2 − 4)/(7 − (−1)) = −6/8 = −3/4
(c) Perpendicular bisector:
Perpendicular gradient: m⊥ = −1 ÷ (−3/4) = 4/3
Passes through midpoint M(3, 1):
y − 1 = (4/3)(x − 3)
y − 1 = (4/3)x − 4
y = (4/3)x − 3
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Q10 — Finding k from gradient condition
(a) Apply the gradient formula to (k, 3) and (2k, −1):
m = (−1 − 3)/(2k − k) = −4/k
Set equal to −2: −4/k = −2
Multiply both sides by k: −4 = −2k
Divide by −2: k = 2
(b) Points are (2, 3) and (4, −1).
Using m = −2 and point (2, 3):
y − 3 = −2(x − 2) → y − 3 = −2x + 4 → y = −2x + 7
Check with (4, −1): y = −8 + 7 = −1 ✓