L49 — Algebra and Coordinate Geometry Review — Solutions

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1. Q1 — Solving quadratics

   Fluency
   • (a) x²−5x+6=0 ⇒ (x−2)(x−3)=0 ⇒ x = 2 or x = 3.
   • (b) 2x²+x−6=0 ⇒ (2x−3)(x+2)=0 ⇒ x = 3/2 or x = −2.
   • (c) x = (6 ± √(36−8))/2 = (6 ± √28)/2 = 3 ± √7.
   • (d) Δ = 16 − 20 = −4 < 0 ⇒ no real solutions.

2. Q2 — Parabola features

   Fluency
   • (a) x = −8/(2×−2) = 2. y = −2(4)+16−3 = 5. Vertex: (2, 5).
   • (b) x = 2.
   • (c) f(0) = −3. y-intercept: (0, −3).
   • (d) Δ = 64 − 4(−2)(−3) = 64 − 24 = 40 > 0. Two x-intercepts.

3. Q3 — Coordinate geometry essentials

   Fluency
   • (a) m = (1−5)/(−4−2) = −4/−6 = 2/3.
   • (b) M = ((2+(−4))/2, (5+1)/2) = (−1, 3).
   • (c) d = √((−4−2)²+(1−5)²) = √(36+16) = √52 = 2√13.
   • (d) y−5 = (2/3)(x−2) ⇒ y = (2/3)x + 11/3. Or: y = (2/3)x + 3⅔.

4. Q4 — Circle equations

   Fluency
   • (a) (x−3)² + (y+2)² = 25.
   • (b) (x−3)²+(y+2)² = 12+9+4 = 25. Centre (3, −2), radius 5.
   • (c) (7−3)²+(−2+2)² = 16 < 25 ⇒ inside the circle.
   • (d) r = √(9+16) = 5. x² + y² = 25.

5. Q5 — Factor and remainder theorems

   Understanding
   • (a) P(−1) = −1+2+5−6 = 0 ✓. (x+1) is a factor.
   • (b) Divide: x³+2x²−5x−6 = (x+1)(x²+x−6) = (x+1)(x+3)(x−2).
   • (c) x = −1, x = −3, x = 2.
   • (d) P(3) = 27+18−15−6 = 24.

6. Q6 — Intersection of a line and quadratic

   Understanding
   • (a) x²−3x+2 = x+3 ⇒ x²−4x−1 = 0.
   • (b) x = (4±√(16+4))/2 = (4±√20)/2 = 2±√5.
   • (c) y = x+3: y = 5+√5 and y = 5−√5. Points: (2+√5, 5+√5) and (2−√5, 5−√5).
   • (d) Δ = 16+4 = 20 > 0 ✓. Two intersection points confirmed.

7. Q7 — Functions: domain, range and notation

   Understanding
   • (a) x − 1 ≥ 0 ⇒ domain: x ≥ 1.
   • (b) f(g(x)) = √(2x+3−1) = √(2x+2) = √(2(x+1)).
   • (c) f(5) = √4 = 2. g(f(5)) = g(2) = 2(2)+3 = 7.
   • (d) y = 2x+3 ⇒ x = (y−3)/2. g⁻¹(x) = (x−3)/2.

8. Q8 — Perpendicular lines and coordinate proofs

   Understanding
   • (a) m_PQ=(0−4)/(3−0)=−4/3. m_QR=(6−0)/(9−3)=1. m_PR=(6−4)/(9−0)=2/9.
   • (b) m_PQ × m_QR = (−4/3)(1) = −4/3 ≠ −1. Check PQ and QR directly: no right angle at Q with these points. Note: right angle is where the product of adjacent side gradients equals −1. No standard right angle at Q with these values — check question intent.
   • (c) Midpoint of PR = ((0+9)/2, (4+6)/2) = (4.5, 5). Perpendicular gradient = −1/m_PR = −9/2. Equation: y−5 = (−9/2)(x−4.5) ⇒ y = −(9/2)x + 25.25.
   • (d) Find perpendicular bisector of PQ (midpoint (1.5,2), gradient 3/4): y−2=(3/4)(x−1.5). Solve with perp bisector of PR. Circumcentre found at intersection of the two perpendicular bisectors.

9. Q9 — Quadratic in context

   Problem Solving
   • (a) Vertex: t=−20/(2×−5)=2 s. h(2)=−20+40+60=maximum height 80 m at t=2 s.
   • (b) −5t²+20t+60=0 ⇒ t²−4t−12=0 ⇒ (t−6)(t+2)=0. t = 6 s (t>0).
   • (c) h(1)=−5+20+60=75 m. h(5)=−125+100+60=35 m.
   • (d) Domain: 0 ≤ t ≤ 6 s. Range: 0 ≤ h ≤ 80 m.

10. Q10 — Extended: quadratic and coordinate geometry

    Problem Solving
    • (a) x-intercepts: (x−1)(x−3)=0 ⇒ (1,0) and (3,0). y-intercept: y=3 ⇒ (0,3).
    • (b) (x−3)² + (y−1)² = 5.
    • (c) Vertex of parabola: x=2, y=4−8+3=−1. So vertex (2,−1). Distance from (3,1): d=√((3−2)²+(1−(−1))²)=√(1+4)=√5.
    • (d) Substitute y=x²−4x+3 into (x−3)²+(y−1)²=5: (x−3)²+(x²−4x+2)²=5. Expanding gives a quartic. The vertex (2,−1) is on the circle (distance = √5 = r) ⇒ the parabola is tangent to the circle at (2,−1) — exactly one point of tangency there.