Distance and Midpoint — Solutions

Click any answer to watch the solution video.

1. Midpoint

   1. (2,4) and (8,10): (5, 7)
   2. (0,0) and (6,−4): (3, −2)
   3. (−3,1) and (5,7): (1, 4)
   4. (1,−5) and (7,3): (4, −1)
   5. (−2,−6) and (4,2): (1, −2)
   6. (3,9) and (−1,−3): (1, 3)
   7. (0,5) and (10,−1): (5, 2)
   8. (−4,−2) and (6,8): (1, 3)

2. Distance (exact)

   1. (0,0) and (3,4): √(9+16) = √25 = 5
   2. (1,2) and (4,6): √(9+16) = √25 = 5
   3. (2,3) and (8,11): √(36+64) = √100 = 10
   4. (−1,1) and (5,9): √(36+64) = √100 = 10
   5. (0,0) and (5,5): √(25+25) = √50 = 5√2
   6. (3,7) and (3,1): √(0+36) = 6 (vertical segment)
   7. (−2,4) and (4,−4): √(36+64) = √100 = 10
   8. (1,−3) and (7,5): √(36+64) = √100 = 10

3. Finding the other endpoint

   1. M(5,3), A(2,1): B_x = 2×5−2 = 8; B_y = 2×3−1 = 5. B = (8, 5)
   2. M(0,4), A(−3,7): B_x = 0−(−3) = 3; B_y = 8−7 = 1. B = (3, 1)
   3. M(6,−1), A(2,3): B_x = 12−2 = 10; B_y = −2−3 = −5. B = (10, −5)
   4. M(−2,5), A(4,2): B_x = −4−4 = −8; B_y = 10−2 = 8. B = (−8, 8)

4. Classify triangles

   1. A(0,0), B(4,0), C(2,3): AB = 4; AC = √(4+9) = √13; BC = √(4+9) = √13. AC = BC. Isosceles.
   2. P(1,1), Q(5,1), R(3, 1+2√3): PQ = 4; PR = √(4+(2√3)²) = √(4+12) = 4; QR = √(4+12) = 4. All sides = 4. Equilateral.
   3. X(−2,−1), Y(2,2), Z(3,−2): XY = √(16+9) = 5; YZ = √(1+16) = √17; XZ = √(25+1) = √26. All different. Scalene.

5. Midpoints in geometry

   1. Rectangle diagonals: Mid AC = ((1+7)/2,(1+5)/2) = (4,3). Mid BD = ((7+1)/2,(1+5)/2) = (4,3). Same midpoint ✓ — diagonals bisect each other.
   2. M(3,2), P(1,k), Q_y=6: (k+6)/2 = 2 ⇒ k+6 = 4 ⇒ k = −2
   3. Right triangle A(0,0), B(6,0), C(6,4): Hypotenuse AC: A(0,0) to C(6,4); midpoint M=(3,2). MA=√(9+4)=√13; MB=√(9+4)=√13; MC=√(9+4)=√13. All equal ✓
   4. Square vertices (0,0),(4,0),(4,4),(0,4); midpoints (2,0),(4,2),(2,4),(0,2): All sides = √(4+4) = 2√2. Adjacent gradients: (0−2)/(4−2)=−1 and (4−0)/(2−4)=... product = −1. A square (rhombus with right angles).

6. Problem solving

   1. Towns A(3,7) and B(11,1):
      1. d = √[(11−3)²+(1−7)²] = √(64+36) = √100 = 10 units = 100 km
      2. M = ((3+11)/2,(7+1)/2) = (7, 4)
   2. Drone L(0,0) → W(9,12) → P(18,0):
      1. LW = √(81+144) = √225 = 15. WP = √(81+144) = √225 = 15. Total = 30 units
      2. LP = √(324+0) = 18 units
      3. Extra distance = 30 − 18 = 12 units
   3. Circle centre C(3,−1), point A(7,2):
      1. r = √[(7−3)²+(2−(−1))²] = √(16+9) = √25 = 5
      2. C is the midpoint of diameter AB. B_x = 2(3)−7 = −1; B_y = 2(−1)−2 = −4. B = (−1, −4)
   4. (5,6) equidistant from (1,3) and (8,9)?: d₁ = √(16+9) = 5. d₂ = √(9+9) = √18 = 3√2 ≈ 4.24. 5 ≠ 3√2. The claim is incorrect.

7. Perimeters of polygons

   1. Triangle A(0,0), B(6,0), C(3,4): AB=6; AC=√(9+16)=5; BC=√(9+16)=5. Perimeter = 6+5+5 = 16 units. (Isosceles.)
   2. Quadrilateral P(1,2), Q(5,2), R(7,6), S(3,6): PQ=4; QR=√(4+16)=√20=2√5; RS=4; SP=√(4+16)=2√5. PQ=RS and QR=SP, all angles not necessarily 90°. Perimeter=8+4√5 ≈ 16.9 units. Parallelogram (opposite sides equal).
   3. Triangle X(−3,1), Y(3,1), Z(0,5): XY=6; XZ=√(9+16)=5; YZ=√(9+16)=5. XZ=YZ ⇒ Isosceles. Perimeter=16.

8. Distance and midpoint in context

   1. Hiker A(2,5), B(14,10), C(8,18):
      1. AB=√(144+25)=√169=13. BC=√(36+64)=√100=10. Total = 23 km
      2. AC=√(36+169)=√205 ≈ 14.3 km
      3. Midpoint of BC = ((14+8)/2,(10+18)/2) = (11, 14)
   2. M(4,7) midpoint of AB, A(1,3):
      1. B_x=2(4)−1=7; B_y=2(7)−3=11. B=(7,11)
      2. AB=√(36+64)=√100=10 units

9. Geometric proofs

   1. Parallelogram A(1,1), B(5,1), C(6,4), D(2,4): Diagonal AC: midpoint=((1+6)/2,(1+4)/2)=(3.5,2.5). Diagonal BD: midpoint=((5+2)/2,(1+4)/2)=(3.5,2.5). Same midpoint ⇒ diagonals bisect each other ⇒ parallelogram. ✓
   2. P(0,0), Q(4,0), R(2,2√3): PQ=4; PR=√(4+12)=√16=4; QR=√(4+12)=4. All sides = 4 ⇒ equilateral. Perimeter = 12 units.
   3. Triangle A(0,0), B(8,0), C(4,6); midpoints of sides: M_AB=(4,0); M_BC=(6,3); M_AC=(2,3). Smaller triangle sides: (4,0)↔(6,3)=√(4+9)=√13; (6,3)↔(2,3)=4; (2,3)↔(4,0)=√(4+9)=√13. Perim=4+2√13. Original: AB=8; BC=√(16+36)=√52=2√13; AC=√(16+36)=2√13. Perim=8+4√13. Half=4+2√13 ✓

10. Finding unknown coordinates

    1. A(3,k), B(0,1), distance 5: √((3−0)²+(k−1)²)=5 ⇒ 9+(k−1)²=25 ⇒ (k−1)²=16 ⇒ k−1=±4 ⇒ k=5 or k=−3
    2. M(5,2), P(p,6), Q(9,q): (p+9)/2=5 ⇒ p=1. (6+q)/2=2 ⇒ q=−2. p=1, q=−2
    3. A(0,0), B(8,0), C equidistant from A and B with AC=5: Equidistant from A and B ⇒ C lies on perpendicular bisector of AB: x=4. AC=√(16+y²)=5 ⇒ 16+y²=25 ⇒ y²=9 ⇒ y=±3. C=(4,3) or C=(4,−3)
    4. Circle O(2,3), T(6,6), midpoint of TU is M(5,5): U_x=2(5)−6=4; U_y=2(5)−6=4. U=(4,4)