L43 Solutions — Distance, Midpoint and Gradient

1. Distance formula. Fluency

   • (a) d = √(3²+4²) = √25 = 5.
   • (b) d = √(3²+4²) = √25 = 5.
   • (c) d = √(6²+8²) = √(36+64) = √100 = 10.
   • (d) 3²+y²=25 ⇒ y²=16 ⇒ y = 4 or y = −4.

2. Midpoint formula. Fluency

   • (a) M = ((2+8)/2, (4+10)/2) = (5, 7).
   • (b) M = ((−3+7)/2, (5−1)/2) = (2, 2).
   • (c) (1+x)/2=4 ⇒ x=7. (7+y)/2=3 ⇒ y=−1. D = (7, −1).
   • (d) (3+x)/2=0 ⇒ x=−3. (4+y)/2=−2 ⇒ y=−8. B = (−3, −8).

3. Gradient. Fluency

   • (a) m = (11−3)/(5−1) = 8/4 = 2.
   • (b) m = (0−4)/(2−0) = −4/2 = −2.
   • (c) Parallel: same gradient = −3.
   • (d) m × 2 = −1 ⇒ m = −1/2.

4. Parallel and perpendicular. Fluency

   • (a) m₁ = 6/2 = 3. m₂ = (10−4)/(3−1) = 6/2 = 3. Same gradient: yes, parallel.
   • (b) m × 5/2 = −1 ⇒ m = −2/5.
   • (c) m(AB) = (4−1)/(3−0) = 1. m(CD) = (3−0)/(5−2) = 1. Same: parallel.
   • (d) m of (0,3)→(4,1) = (1−3)/(4−0) = −1/2. Perpendicular: m = 2.

5. Triangle on the coordinate plane. Understanding

   • (a) AB = √(36+0) = 6. BC = √(9+16) = √25 = 5. CA = √(9+16) = 5.
   • (b) Mid(AB) = (3,0). Mid(BC) = (4.5,2). Mid(CA) = (1.5,2).
   • (c) BC = CA = 5. Triangle is isosceles (two equal sides) ✓.
   • (d) m(AB) = 0 (horizontal). Altitude from C is vertical (x=3), gradient undefined. Horizontal ⊥ vertical, so they are perpendicular ✓.

6. Collinearity and section formula. Understanding

   • (a) m(AB)=(5−1)/(3−1)=2. m(BC)=(9−5)/(5−3)=2. Equal gradients ⇒ collinear ✓.
   • (b) rise = gradient × run = 3 × 4 = 12.
   • (c) Ratio 2:1 from A: P = A + 2/3(B−A) = (0+4, 0+6) = (4, 6).
   • (d) Gradient = 5/2. Length = √(2²+5²) = √29 ≈ 5.39 m.

7. Applications of midpoint. Understanding

   • (a) Diag AC: mid = ((0+6)/2,(0+4)/2)=(3,2). Diag BD: mid = ((6+0)/2,(0+4)/2)=(3,2). Same midpoint (3,2) ⇒ diagonals bisect each other ✓.
   • (b) Mid(QR) = (4,0). Median from P: d=√((5−4)²+(7−0)²)=√(1+49)=√50=5√2.
   • (c) (k+2)/2=4 ⇒ k=6. Check y: (3+k)/2=5 ⇒ k=7. Contradiction — need both to hold: recheck. x: (k+2)/2=4 ⇒ k=6. y: (3+k)/2=5 ⇒ k=7. There is no single k satisfying both. (Note: this is a trick question — for a valid midpoint, we need A=(k,3) and B=(2,k) giving M=((k+2)/2,(3+k)/2)=(4,5). So (k+2)/2=4 ⇒ k=6 and (3+k)/2=5 ⇒ k=7. Both must hold — since they give different values, no solution exists for a single k.)
   • (d) Each ship’s current position is the midpoint of the other’s 0-and-10 endpoints. By symmetry they converge at (5, 0).

8. Gradient and collinearity. Understanding

   • (a) Rise = m × run = 4 × 3 = 12.
   • (b) m(AB)=(5−2)/(t−0)=3/t. m(AC)=(14−2)/(8−0)=12/8=3/2. Set equal: 3/t=3/2 ⇒ t=2.
   • (c) m(AB)=(3−1)/(5−1)=2/4=1/2. Adjacent (perpendicular): m=−2.
   • (d) m of (0,2)→(3,4) = 2/3. Perpendicular through (3,4): m = −3/2.

9. Coordinate geometry in context. Problem Solving

   • (a) AB=8, BC=6, CD=8, DA=6 (rectangle). Perimeter = 28 km.
   • (b) Diagonals: AC from (0,0) to (8,6), BD from (8,0) to (0,6). Midpoint of AC = (4,3). Midpoint of BD = (4,3). School at (4, 3).
   • (c) Length = √(8²+6²) = √100 = 10 km. Gradient = 6/8 = 3/4.
   • (d) Beta(8,0)→Delta(0,6): m = (6−0)/(0−8) = −3/4. Product of gradients: 3/4 × (−3/4) = −9/16 ≠ −1. Actually these are not perpendicular. The diagonals of a rectangle ARE perpendicular only for a square. Here 3/4 × (−3/4) = −9/16 ≠ −1, confirming these diagonals are NOT perpendicular (the rectangle is 8×6, not a square). Note to student: AB and BD refer to different lines — the two routes are Alpha→Gamma and Beta→Delta (the diagonals).

10. Triangle geometry. Problem Solving

    • (a) AC=√((5−2)²+(7−1)²)=√(9+36)=√45=3√5. BC=√((5−8)²+(7−1)²)=√(9+36)=3√5. Equal ⇒ isosceles ✓. AB=8−2=6.
    • (b) G = ((2+8+5)/3, (1+1+7)/3) = (15/3, 9/3) = (5, 3).
    • (c) Mid(AB)=(5,1). AB is horizontal (m=0), so perpendicular bisector is vertical: x = 5.
    • (d) C=(5,7): x-coordinate is 5 ✓. Perpendicular bisector of AB is x=5, and C lies on it. Since C is equidistant from A and B (both distances equal 3√5), the triangle must be isosceles ✓.