FORMULA CHAPTER
Distance (AB) = \sqrt{(\mathrm{x}_2-\mathrm{x}_1)^2+(\mathrm{y}_2-\mathrm{y}_1)^2}
AB + BC = AC
AB = AC
AB = AC and AB² + AC² = BC²
AB = BC = CA
AB = BC = CD = DA, AC = BD, AC² = 2AB²
\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\quad\mathrm{and}\quad \mathrm{AC}\ne\mathrm{BD}
AB = CD, BC = DA, AC = BD, AC² = AB² + BC²
PA = PB
Distance (AB) = \sqrt{(\mathrm{x}_2-\mathrm{x}_1)^2+(\mathrm{y}_2-\mathrm{y}_1)^2}
Question: Find the distance between the points A(2, 3) and B(8, 11).
Calculation: AB = √[(8 − 2)² + (11 − 3)²]
AB = √(6² + 8²)
⇒ AB = √(36 + 64)
⇒ AB = √100
⇒ AB = 10
Answer: The distance between A and B is 10 units.
AB + BC = AC
Question: Show that A(0, 0), B(3, 4) and C(6, 8) are collinear.
Calculation:
Step 1: AB = √[(3 − 0)² + (4 − 0)²] = √25 = 5 units.
Step 2: BC = √[(6 − 3)² + (8 − 4)²] = √25 = 5 units.
Step 3: AC = √[(6 − 0)² + (8 − 0)²] = √100 = 10 units.
Step 4: AB + BC = 5 + 5 = 10 = AC.
Answer: A, B and C are collinear.
AB = AC
Question: Prove that the triangle with vertices A(0, 4), B(−3, 0) and C(3, 0) is isosceles.
Calculation:
Step 1: AB² = (−3 − 0)² + (0 − 4)² = 9 + 16 = 25.
Step 2: AC² = (3 − 0)² + (0 − 4)² = 9 + 16 = 25.
Step 3: AB² = AC², therefore AB = AC = 5 units.
Answer: △ABC is an isosceles triangle, with equal sides AB and AC.
AB = AC and AB² + AC² = BC²
Question: Prove that A(0, 0), B(4, 0) and C(0, 4) form a right isosceles triangle.
Calculation:
Step 1: AB² = (4 − 0)² + (0 − 0)² = 16.
Step 2: AC² = (0 − 0)² + (4 − 0)² = 16.
Step 3: BC² = (0 − 4)² + (4 − 0)² = 16 + 16 = 32.
Step 4: AB = AC and AB² + AC² = 16 + 16 = 32 = BC².
Answer: △ABC is a right isosceles triangle, right-angled at A.
AB = BC = CA
Question: Prove that A(0, 0), B(4, 0) and C(2, 2√3) form an equilateral triangle.
Calculation:
Step 1: AB² = (4 − 0)² + (0 − 0)² = 16.
Step 2: BC² = (2 − 4)² + (2√3 − 0)² = 4 + 12 = 16.
Step 3: CA² = (0 − 2)² + (0 − 2√3)² = 4 + 12 = 16.
Step 4: AB = BC = CA = 4 units.
Answer: △ABC is an equilateral triangle.
AB = BC = CD = DA, AC = BD, AC² = 2AB²
Question: Prove that A(0, 0), B(4, 0), C(4, 4) and D(0, 4) form a square.
Calculation:
Step 1: AB² = BC² = CD² = DA² = 16, so all four sides are 4 units.
Step 2: AC² = (4 − 0)² + (4 − 0)² = 32.
Step 3: BD² = (0 − 4)² + (4 − 0)² = 32, so AC = BD.
Step 4: AC² = 32 = 2 × 16 = 2AB².
Answer: ABCD is a square.
\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\quad\mathrm{and}\quad \mathrm{AC}\ne\mathrm{BD}
Question: Prove that A(0, 2), B(3, 0), C(6, 2) and D(3, 4) form a rhombus.
Calculation:
Step 1: AB² = (3 − 0)² + (0 − 2)² = 9 + 4 = 13.
Step 2: BC² = CD² = DA² = 13, so all four sides are equal.
Step 3: AC = 6 units and BD = 4 units, so AC ≠ BD.
Answer: ABCD is a rhombus and not a square.
AB = CD, BC = DA, AC = BD, AC² = AB² + BC²
Question: Prove that A(0, 0), B(6, 0), C(6, 3) and D(0, 3) form a rectangle.
Calculation:
Step 1: AB = CD = 6 units and BC = DA = 3 units.
Step 2: AC² = (6 − 0)² + (3 − 0)² = 36 + 9 = 45.
Step 3: BD² = (0 − 6)² + (3 − 0)² = 36 + 9 = 45, so AC = BD.
Step 4: AB² + BC² = 36 + 9 = 45 = AC².
Answer: ABCD is a rectangle.
PA = PB
Question: Show that P(3, 2) is equidistant from A(1, 0) and B(5, 4).
Calculation:
Step 1: Find PA²:
PA² = (3 − 1)² + (2 − 0)² = 4 + 4 = 8
Step 2: Find PB²:
PB² = (3 − 5)² + (2 − 4)² = 4 + 4 = 8
Step 3: Compare the distances:
PA² = PB², so PA = PB = √8 = 2√2 coordinate units.
Answer: P is equidistant from A and B.
