L44 Solutions — Equations of Lines

1. Gradient–intercept form. Fluency

   • (a) y = 4x − 3.
   • (b) Gradient = −3, y-intercept = 7.
   • (c) Set y = 0: 0 = 2x − 6 ⇒ x = 3.
   • (d) y = 5(−1) + 2 = −5 + 2 = −3.

2. Point–gradient form. Fluency

   • (a) y − 5 = 2(x − 3) ⇒ y = 2x − 1. y = 2x − 1.
   • (b) m = −1, through (0,4): y = −x + 4.
   • (c) y − 1 = ½(x + 4) ⇒ y − 1 = x/2 + 2 ⇒ y = x/2 + 3.
   • (d) y − 3 = 4(x + 1) ⇒ y − 3 = 4x + 4 ⇒ y = 4x + 7.

3. Two-point form. Fluency

   • (a) m = (7−3)/(2−0) = 2. b = 3. y = 2x + 3.
   • (b) m = (5−(−1))/(4−1) = 6/3 = 2. y + 1 = 2(x − 1) ⇒ y = 2x − 3.
   • (c) m = (6−0)/(0−3) = −2. y = −2x + 6.
   • (d) m = (5−(−4))/(2−(−1)) = 9/3 = 3. y − 5 = 3(x − 2) ⇒ y = 3x − 1 ⇒ 3x − y − 1 = 0.

4. Intersection of lines. Fluency

   • (a) x + 2 = −x + 6 ⇒ 2x = 4 ⇒ x = 2, y = 4. (2, 4).
   • (b) 3x − 1 = x + 3 ⇒ 2x = 4 ⇒ x = 2, y = 5. (2, 5).
   • (c) Add equations: 3x = 9 ⇒ x = 3. y = 3 − 1 = 2. (3, 2).
   • (d) Both have gradient 2 but different y-intercepts (1 and −3). No — they are parallel and never intersect.

5. Lines on a graph. Understanding

   • (a) A: y = x. B: y = −x + 4. C: y = 2.
   • (b) A: 2 = 2 ✓. B: −2 + 4 = 2 ✓. C: 2 = 2 ✓. All three pass through (2, 2).
   • (c) m_A = 1, m_B = −1. Product = −1 ⇒ A and B are perpendicular. (C is horizontal, perpendicular to a vertical line only.)
   • (d) y − 2 = 3(x − 2) ⇒ y = 3x − 4.

6. Parallel and perpendicular through a point. Understanding

   • (a) Parallel: m = 4. y − 3 = 4(x − 1) ⇒ y = 4x − 1.
   • (b) Perpendicular to m = −3: m = 1/3. Through (0,2): y = x/3 + 2.
   • (c) Parallel: k = 3. Perpendicular: k × 3 = −1 ⇒ k = −1/3.
   • (d) 3x − 4y = 12 ⇒ y = 3x/4 − 3, so m = 3/4. Perpendicular: m = −4/3. Through (3,−1): y + 1 = −4/3(x − 3) ⇒ y = −4x/3 + 3.

7. Lines and regions. Understanding

   • (a) At x = 3: y = 2(3) − 1 = 5. The point (3, 5) gives exactly y = 5 on the line. The point lies on the line (neither above nor below).
   • (b) Points above y = x + 3 satisfy y > x + 3.
   • (c) The two lines are parallel (same gradient 1). At any x, the vertical gap = (x + 4) − x = 4 units. At x = 2: gap = (6) − (2) = 4.
   • (d) 2x + 3y = 12 ⇒ y = −2x/3 + 4. Gradient = −2/3. x-intercept (y=0): x = 6. y-intercept (x=0): y = 4.

8. Systems of linear equations. Understanding

   • (a) L1∩L2: x+1=2x−1 ⇒ x=2,y=3 ⇒ (2,3). L1∩L3: x+1=−x+3 ⇒ x=1,y=2 ⇒ (1,2). L2∩L3: 2x−1=−x+3 ⇒ x=4/3,y=5/3 ⇒ (4/3, 5/3).
   • (b) Substitute (1,5) into L₁: 5 = 2(1) + c ⇒ c = 3. Verify L₂: y = −1 + 6 = 5 ✓.
   • (c) m = (0−(−6))/(4−0) = 6/4 = 3/2. Equation: y = 3x/2 − 6.
   • (d) Rearrange: 6x − 2y + 10 = 0 ⇒ y = 3x + 5. Both lines have gradient 3, so they are parallel for all k. For distinct parallel lines: k ≠ 5 (k = 5 gives the same line).

9. Applied linear modelling. Problem Solving

   • (a) C₁(h) = 60h + 80. C₂(h) = 50h.
   • (b) 60h + 80 = 50h ⇒ 10h = −80 ⇒ h = −8. No positive solution — Plumber 2 is always cheaper (higher rate and call-out fee for Plumber 1 means the lines never cross for h > 0).
   • (c) C₁ starts at (0, 80) with steeper slope. C₂ starts at origin with shallower slope. C₁ is always above C₂ for h > 0 — lines diverge rather than intersect.
   • (d) h = 5: C₁ = 300 + 80 = $380. C₂ = $250. Plumber 2 cheaper. h = 1: C₁ = $140. C₂ = $50. Plumber 2 cheaper for both jobs.

10. Perpendicular bisector and triangle properties. Problem Solving

    • (a) m(AB) = (2−8)/(8−2) = −6/6 = −1. Perpendicular gradient = 1.
    • (b) Midpoint M = ((2+8)/2, (8+2)/2) = (5, 5). Perpendicular bisector: y − 5 = 1(x − 5) ⇒ y = x.
    • (c) On y = x with x = 3: y = 3. C = (3, 3).
    • (d) CA = √((3−2)²+(3−8)²) = √(1+25) = √26. CB = √((3−8)²+(3−2)²) = √(25+1) = √26. CA = CB = √26 ✓.