Geometry, Trig & Calculus

Coordinate Geometry and Mensuration

SSS 2 · SSS 3

✓By the end of this lesson students will be able to calculate the distance between two points and the gradient of a straight line.
✓By the end of this lesson students will be able to determine the equation of a straight line in various forms and solve problems involving parallel and perpendicular lines.
✓By the end of this lesson students will be able to find the equation of a circle given its centre and radius, or other relevant information.
✓By the end of this lesson students will be able to calculate the perimeter and area of plane shapes, and the surface area and volume of common solids.
✓By the end of this lesson students will be able to apply coordinate geometry and mensuration principles to solve real-world problems.

Key concepts

Distance Between Two Points

The distance between two points P(x₁, y₁) and Q(x₂, y₂) in a Cartesian coordinate system is found using the distance formula, which is derived from the Pythagorean theorem.

d = √((x₂ - x₁)² + (y₂ - y₁)²)

Gradient of a Straight Line

The gradient (or slope) of a straight line measures its steepness. It is the ratio of the change in the y-coordinates to the change in the x-coordinates between any two points on the line.

m = (y₂ - y₁) / (x₂ - x₁)

Equation of a Straight Line

The equation of a straight line can be expressed in several forms:\n1. Gradient-intercept form: y = mx + c, where 'm' is the gradient and 'c' is the y-intercept.\n2. Point-gradient form: y - y₁ = m(x - x₁), where 'm' is the gradient and (x₁, y₁) is a point on the line.\n3. Two-point form: (y - y₁) / (x - x₁) = (y₂ - y₁) / (x₂ - x₁), where (x₁, y₁) and (x₂, y₂) are two points on the line.

y = mx + c OR y - y₁ = m(x - x₁)

Parallel and Perpendicular Lines

Two lines are parallel if they have the same gradient (m₁ = m₂). Two lines are perpendicular if the product of their gradients is -1 (m₁ × m₂ = -1). This implies that m₂ = -1/m₁.

Parallel: m₁ = m₂\nPerpendicular: m₁m₂ = -1

Equation of a Circle

The equation of a circle defines all points (x, y) that are a fixed distance (radius, r) from a fixed point (centre).\n1. Centre at the origin (0, 0): x² + y² = r².\n2. Centre at (a, b): (x - a)² + (y - b)² = r².\n3. General form: x² + y² + 2gx + 2fy + c = 0, where the centre is (-g, -f) and the radius r = √(g² + f² - c).

(x - a)² + (y - b)² = r²

Mensuration of Plane Shapes

Mensuration involves calculating lengths, areas, and volumes of geometric figures.\n\nCommon Plane Shapes:\n* **Rectangle**: Perimeter = 2(l + w), Area = l × w\n* **Square**: Perimeter = 4l, Area = l²\n* **Triangle**: Area = ½ × base × height\n* **Parallelogram**: Area = base × height\n* **Trapezium**: Area = ½(a + b)h, where 'a' and 'b' are parallel sides\n* **Circle**: Circumference = 2πr or πd, Area = πr²\n* **Sector of a Circle**: Arc length = (θ/360°) × 2πr, Area = (θ/360°) × πr² (where θ is in degrees)

Varies by shape

Mensuration of Solid Shapes

Formulas for surface area and volume of common three-dimensional solids:\n* **Cuboid**: Volume = l × w × h, Surface Area = 2(lw + lh + wh)\n* **Cube**: Volume = l³, Surface Area = 6l²\n* **Cylinder**: Volume = πr²h, Curved Surface Area = 2πrh, Total Surface Area = 2πr(r + h)\n* **Cone**: Volume = ⅓πr²h, Curved Surface Area = πrl (where l is slant height), Total Surface Area = πr(r + l)\n* **Pyramid**: Volume = ⅓ × Base Area × height\n* **Sphere**: Volume = ⅔πr³, Surface Area = 4πr²\n* **Hemisphere**: Volume = ⅑πr³, Curved Surface Area = 2πr², Total Surface Area = 3πr²

Varies by shape

Key facts to remember

1Distance formula: d = √((x₂ - x₁)² + (y₂ - y₁)²)
2Gradient formula: m = (y₂ - y₁) / (x₂ - x₁)
3Equation of a straight line: y = mx + c or y - y₁ = m(x - x₁)
4Parallel lines have equal gradients (m₁ = m₂).
5Perpendicular lines have gradients whose product is -1 (m₁m₂ = -1).
6Equation of a circle with centre (a, b) and radius r: (x - a)² + (y - b)² = r².
7Area of a circle = πr², Circumference = 2πr.
8Volume of a cylinder = πr²h, Volume of a cone = ⅓πr²h, Volume of a sphere = ⅔πr³.

Worked examples

Example 1

Given points A(2, 3) and B(5, -1):\n(a) Find the gradient of the line AB.\n(b) Find the equation of the line AB.\n(c) Find the equation of a line perpendicular to AB and passing through the point C(1, 4).

I(a) To find the gradient of line AB, use the formula m = (y₂ - y₁) / (x₂ - x₁).\nLet (x₁, y₁) = (2, 3) and (x₂, y₂) = (5, -1).\nm₁ = (-1 - 3) / (5 - 2)\nm₁ = -4 / 3

II(b) To find the equation of line AB, use the point-gradient form y - y₁ = m(x - x₁).\nUsing point A(2, 3) and gradient m₁ = -4/3:\ny - 3 = (-4/3)(x - 2)\nMultiply by 3 to clear the fraction:\n3(y - 3) = -4(x - 2)\n3y - 9 = -4x + 8\nRearrange into general form:\n4x + 3y - 9 - 8 = 0\n4x + 3y - 17 = 0

III(c) For a line perpendicular to AB, the product of their gradients must be -1.\nLet m₂ be the gradient of the perpendicular line.\nm₁ × m₂ = -1\n(-4/3) × m₂ = -1\nm₂ = -1 × (-3/4)\nm₂ = 3/4\nNow, use the point-gradient form with point C(1, 4) and gradient m₂ = 3/4:\ny - 4 = (3/4)(x - 1)\nMultiply by 4:\n4(y - 4) = 3(x - 1)\n4y - 16 = 3x - 3\nRearrange into general form:\n3x - 4y - 3 + 16 = 0\n3x - 4y + 13 = 0

Answer

(a) Gradient of AB = -4/3\n(b) Equation of line AB: 4x + 3y - 17 = 0\n(c) Equation of the perpendicular line: 3x - 4y + 13 = 0

Always simplify your final equation to the general form Ax + By + C = 0, where A, B, C are integers and A is usually positive.

Example 2

A circle has its centre at (3, -2) and passes through the point (7, 1).\n(a) Find the radius of the circle.\n(b) Find the equation of the circle.

I(a) The radius 'r' is the distance between the centre (3, -2) and the point (7, 1) on the circumference.\nUsing the distance formula d = √((x₂ - x₁)² + (y₂ - y₁)²):\nr = √((7 - 3)² + (1 - (-2))²)\nr = √((4)² + (3)²)\nr = √(16 + 9)\nr = √25\nr = 5 units

II(b) The equation of a circle with centre (a, b) and radius r is (x - a)² + (y - b)² = r².\nGiven centre (a, b) = (3, -2) and radius r = 5.\n(x - 3)² + (y - (-2))² = 5²\n(x - 3)² + (y + 2)² = 25\nExpanding this to the general form (optional, but good practice):\nx² - 6x + 9 + y² + 4y + 4 = 25\nx² + y² - 6x + 4y + 13 - 25 = 0\nx² + y² - 6x + 4y - 12 = 0

Answer

(a) Radius = 5 units\n(b) Equation of the circle: (x - 3)² + (y + 2)² = 25 OR x² + y² - 6x + 4y - 12 = 0

Both forms of the circle equation are acceptable unless a specific form is requested. The expanded general form is often useful for finding the centre and radius if not directly given.

Example 3

A solid metal cone has a base radius of 7 cm and a perpendicular height of 24 cm. It is melted down and recast into a sphere.\n(a) Calculate the volume of the cone. (Take π = 22/7)\n(b) Calculate the radius of the sphere.

I(a) Volume of a cone = ⅓πr²h\nGiven r = 7 cm, h = 24 cm, π = 22/7.\nVolume = ⅓ × (22/7) × (7)² × 24\nVolume = ⅓ × (22/7) × 49 × 24\nVolume = 22 × 7 × 8 (since 49/7 = 7 and 24/3 = 8)\nVolume = 154 × 8\nVolume = 1232 cm³

II(b) When the cone is melted and recast into a sphere, the volume remains the same.\nVolume of sphere = Volume of cone = 1232 cm³.\nVolume of sphere = ⅔πR³, where R is the radius of the sphere.\n⅔ × (22/7) × R³ = 1232\n(88/21) × R³ = 1232\nR³ = 1232 × (21/88)\nR³ = (1232 × 21) / 88\nR³ = 14 × 21 (since 1232 / 88 = 14)\nR³ = 294\nR = ∛294\nR ≈ 6.65 cm (to 3 significant figures)

Answer

(a) Volume of the cone = 1232 cm³\n(b) Radius of the sphere ≈ 6.65 cm

Remember that when a solid is melted and recast, its volume remains constant. Pay attention to units and the required level of accuracy for answers.

Common mistakes

✗Confusing the signs in the distance formula or gradient formula, especially with negative coordinates.
✗Incorrectly applying the perpendicular gradient rule (e.g., using m₂ = 1/m₁ instead of m₂ = -1/m₁).
✗Expanding (x - a)² as x² - a² instead of x² - 2ax + a² when dealing with circle equations.
✗Using incorrect units or forgetting to include units in final answers for mensuration problems.
✗Mixing up formulas for area and volume, or for different shapes (e.g., using cylinder volume for a cone).

Exam tips

★Always draw a sketch for coordinate geometry problems; it helps visualise the points, lines, and circles.
★Memorise all standard formulas for distance, gradient, equations of lines and circles, and mensuration of common shapes.
★Show all your working steps clearly, as marks are often awarded for method even if the final answer is incorrect.
★Pay close attention to the value of π given in the question (e.g., 22/7 or 3.142) and the required degree of accuracy for your final answer.

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