Module 3: Area of Quadrilaterals
|
Shape |
Formula |
Example |
|
Square |
( \text{Area} = s^2 ) |
Side = 5 cm → Area = 5² = 25 cm² |
|
Rectangle |
( \text{Area} = l × b ) |
Length = 8 cm, Breadth = 5 cm → Area = 8 × 5 = 40 cm² |
|
Parallelogram |
( \text{Area} = b × h ) |
Base = 10 cm, height = 6 cm → Area = 60 cm² |
|
Trapezium |
( \text{Area} = \frac{1}{2}(a + b)h ) |
Bases = 6 cm, 10 cm; height = 4 cm → Area = 32 cm² |
Module 4: Area of a Circle
Formula:
\text{Area} = πr^2
Example:
Circle with radius 7 cm:
\text{Area} = 22/7 × 7 × 7 = 154 \text{ cm²}
- Example: (3×2+2x+5)+(x2−3x+4)=4×2−x+9(3x^2 + 2x + 5) + (x^2 – 3x + 4) = 4x^2 – x + 9.
- Subtraction: Subtract like terms.
- Example: (5×2+3x+2)−(2×2−x+1)=3×2+4x+1(5x^2 + 3x + 2) – (2x^2 – x + 1) = 3x^2 + 4x + 1
- Multiplication: Multiply each term in one polynomial by each term in the other.
- Example: (2x+3)(x+1)=2×2+5x+3(2x + 3)(x + 1) = 2x^2 + 5x + 3.
- Real-Life Application:
- Engineering: Polynomials are used in various fields of engineering, such as modeling physical systems or structures.
- Finance: In business, polynomials can represent compound interest formulas or profit functions.
Topic: Gradient and Distance Between Two Points
Learning Objectives
By the end of this lesson, students should be able to:
Define gradient (slope) and explain its geometric meaning.
Calculate the gradient between two points.
Derive and use the distance formula between two points.
Apply gradient and distance concepts to solve coordinate geometry problems.
Lesson Content
1. Meaning of Gradient
The gradient (or slope) of a line measures how steep the line is.
It shows the change in the y-coordinate for a change in the x-coordinate.
If two points on a line are ( A(x_1, y_1) ) and ( B(x_2, y_2) ),
then the gradient ( m ) is given by:
[
m = \frac{y_2 – y_1}{x_2 – x_1}
]
Example 1:
Find the gradient of the line passing through the points ( (2, 3) ) and ( (6, 7) ).
[
m = \frac{7 – 3}{6 – 2} = \frac{4}{4} = 1
]
So, the gradient of the line is 1.
2. Special Cases
Horizontal Line: ( y_2 – y_1 = 0 \Rightarrow m = 0 )
Vertical Line: ( x_2 – x_1 = 0 \Rightarrow m ) is undefined
3. Meaning of Distance Between Two Points
The distance between two points in the coordinate plane can be found using the distance formula:
[
d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}
]
This formula is derived from the Pythagoras theorem.
Example 2:
Find the distance between the points ( A(3, 4) ) and ( B(7, 1) ).
[
d = \sqrt{(7 – 3)^2 + (1 – 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
]
So, the distance between A and B is 5 units.
4. Relationship Between Gradient and Angle of Inclination
The gradient ( m ) of a line is related to the angle of inclination ( \theta ) by:
[
m = \tan(\theta)
]
If ( m = 1 ), then ( \theta = 45^\circ ).
5. Example (WAEC/IGCSE style)
Find the gradient of the line joining ( P(2, -1) ) and ( Q(8, 5) ),
and the distance between P and Q.
Solution:
[
m = \frac{5 – (-1)}{8 – 2} = \frac{6}{6} = 1
]
[
d = \sqrt{(8 – 2)^2 + (5 – (-1))^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}
]
Classwork / Practice Exercises
Find the gradient of the line joining the points (1, 2) and (5, 8).
Determine the distance between (3, 7) and (9, 4).
What is the gradient of a line parallel to ( y = 2x + 3 )?
Calculate the distance between (–4, –3) and (2, 9).
If a line has a gradient of 3, find the angle of inclination.
Assignment
Find the gradient and distance between points (–2, 1) and (4, 5).
Determine whether the line joining (3, 2) and (9, 8) is parallel to the line joining (1, 1) and (4, 4).
A line passes through A(–5, 2) and B(1, k). If the gradient is ( \frac{1}{2} ), find the value of k.
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What is a Fractional Equation?
A fractional equation is an equation that contains fractions with variables in the denominator.
For example:
\frac{x + 2}{3} = \frac{2x – 1}{5}
or
The goal is to find the value(s) of the variable that make the equation true.
Steps for Solving Fractional Equations
- Find the denominators — list all the denominators in the equation.
- Find the LCM (Lowest Common Multiple) of all the denominators.
- Multiply through by the LCM to eliminate fractions.
- Simplify the resulting equation (it should now be free of denominators).
- Solve the equation as a normal algebraic equation.
- Check for restrictions — values that make any denominator zero are not allowed (they make the equation undefined).
Example 1
Solve:
\frac{x + 1}{2} = \frac{3x – 5}{4}
Step 1: Denominators are 2 and 4.
LCM = 4
Step 2: Multiply through by 4:
4 \times \frac{x + 1}{2} = 4 \times \frac{3x – 5}{4}
What is a Quadratic Equation?
A quadratic equation is an equation in which the highest power of the variable is 2.
It has the general form:
ax^2 + bx + c = 0
where
- (a, b, c) are real numbers, and
- (a \neq 0).
Examples
- (x^2 – 5x + 6 = 0)
- (2x^2 + 3x – 5 = 0)
- (3x^2 = 12x – 9)
Methods of Solving Quadratic Equations
There are 4 main methods:
Factorization Method
Example:
x^2 – 5x + 6 = 0 [/katex]
Find two numbers that multiply to 6 and add to -5 → (-2) and (-3)
x^2 – 2x – 3x + 6 = 0
x(x – 2) – 3(x – 2) = 0
(x – 2)(x – 3) = 0
x = 2 or x = 3
Completing the Square
Example:
x^2 + 6x + 5 = 0
Move the constant to the right:
x^2 + 6x = -5
Take half of the coefficient of (x), square it:
\left(\frac{6}{2}\right)^2 = 9
Add 9 to both sides:
[katext]
x^2 + 6x + 9 = -5 + 9
[/katex]
(x + 3)^2 = 4
x + 3 = \pm 2
x = -3 \pm 2
x = -1 or x = -5
Using the Quadratic Formula
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}Example:
2x^2 + 3x – 2 = 0
Here, (a = 2), (b = 3), (c = -2)
x = \frac{-3 \pm \sqrt{3^2 – 4(2)(-2)}}{2(2)}
x = \frac{-3 \pm \sqrt{9 + 16}}{4}
x = \frac{-3 \pm 5}{4}
<strong>x = (\frac{1}{2})</strong> or <strong>x = -2
Overview
Coordinate geometry, also called analytic geometry, is the study of geometric figures using a coordinate system. It helps us represent points, lines, and shapes on the Cartesian plane using algebraic equations.
Module 1: The Cartesian Plane
Lesson Content
The Cartesian plane consists of two perpendicular lines called axes:
– The x-axis (horizontal)
– The y-axis (vertical)
These axes intersect at a point called the origin (0, 0). Each point on the plane is represented as an ordered pair (x, y).
Example
Plot the points A(2, 3), B(–1, 2), and C(0, –3) on the Cartesian plane.
– A is 2 units right and 3 units up.
– B is 1 unit left and 2 units up.
– C is on the y-axis, 3 units below the origin.
Quick Quiz (MCQs)
- The point (0, 0) is called the:
A. Midpoint
B. Origin
C. Centre
D. Pole
Answer: B - The x-axis and y-axis divide the plane into how many quadrants?
A. Two
B. Three
C. Four
D. Five
Answer: C
Assignment
Identify the quadrant in which the following points lie:
(a) (3, 2)
(b) (–4, 1)
(c) (–2, –3)
(d) (5, –1)
Module 2: Distance Between Two Points
Lesson Content
The distance (d) between two points A(x₁, y₁) and B(x₂, y₂) is given by:
d = √((x₂ – x₁)² + (y₂ – y₁)²)
Example
Find the distance between P(2, 3) and Q(7, 11):
d = √((7 – 2)² + (11 – 3)²) = √(25 + 64) = √89
Quick Quiz (MCQs)
- The distance between (1, 2) and (4, 6) is:
A. 4
B. 5
C. √20
D. √25
Answer: B
Assignment
Find the distance between the following pairs of points:
(a) (2, 5) and (6, 9)
(b) (–1, –3) and (4, 2)
Module 3: Midpoint of a Line Segment
Lesson Content
The midpoint (M) of a line joining A(x₁, y₁) and B(x₂, y₂) is:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Example
Find the midpoint of A(2, 3) and B(6, 11):
M = ((2 + 6)/2, (3 + 11)/2) = (4, 7)
Quick Quiz (MCQs)
- The midpoint of (2, 4) and (6, 8) is:
A. (8, 12)
B. (2, 8)
C. (4, 6)
D. (3, 5)
Answer: C
Assignment
Find the midpoint between (–3, 5) and (7, –9).
Module 4: Gradient (Slope) of a Line
Lesson Content
The gradient (m) of a line passing through points A(x₁, y₁) and B(x₂, y₂) is:
m = (y₂ – y₁)/(x₂ – x₁)
If the line rises as it moves right, the slope is positive.
If it falls as it moves right, the slope is negative.
Example
Find the gradient of a line through (1, 2) and (4, 8):
m = (8 – 2)/(4 – 1) = 6/3 = 2
Quick Quiz (MCQs)
- The gradient of the line joining (2, 3) and (5, 9) is:
A. 1
B. 2
C. 3
D. 4
Answer: B
Assignment
Find the slope of the line passing through (–2, –1) and (3, 4).
Module 5: Equation of a Straight Line
Lesson Content
The general form of a straight-line equation is:
y = mx + c
where:
– m = gradient
– c = y-intercept
If the line passes through (x₁, y₁) with slope m, then:
y – y₁ = m(x – x₁)
Example
Find the equation of a line passing through (2, 3) with gradient 2:
y – 3 = 2(x – 2) ⇒ y = 2x – 1
Quick Quiz (MCQs)
- The line passing through (0, 4) with slope 3 is:
A. y = 3x + 4
B. y = 3x – 4
C. y = 4x + 3
D. y = 4x – 3
Answer: A
Assignment
Find the equation of a line passing through (1, 2) and (3, 6).
Module 6: Parallel and Perpendicular Lines
Lesson Content
Two lines are parallel if they have equal gradients.
Two lines are perpendicular if the product of their gradients = –1.
Example
If line 1 has slope 2, then:
– Any parallel line has slope 2.
– Any perpendicular line has slope –½.
Quick Quiz (MCQs)
- Which of the following lines is perpendicular to y = 2x + 3?
A. y = 2x – 4
B. y = –½x + 1
C. y = ½x – 1
D. y = –2x + 4
Answer: B
Assignment
Find the equation of a line perpendicular to y = 3x + 2 that passes through (0, 4).
– Solve real-life and WAEC/IGCSE-style problems involving the midpoint of a line.
1. Meaning of Midpoint
The midpoint of a line segment is the point that divides the line into two equal halves. If you have two points A(x₁, y₁) and B(x₂, y₂), the midpoint M is given by the formula:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
This means the coordinates of the midpoint are the averages of the x-coordinates and y-coordinates of the two endpoints.
Example 1
Find the midpoint of the line joining A(2, 3) and B(6, 7):
M = ((2 + 6)/2, (3 + 7)/2) = (4, 5)
So, the midpoint is (4, 5).
Example 2
Find the coordinates of the midpoint between P(–4, 2) and Q(8, 10):
M = ((–4 + 8)/2, (2 + 10)/2) = (2, 6)
Therefore, the midpoint of PQ is (2, 6).
Example 3 (Reverse Problem)
If the midpoint of a line joining A(2, 3) and B(x, 7) is M(4, 5), find x.
(4, 5) = ((2 + x)/2, (3 + 7)/2)
Equating x-coordinates:
4 = (2 + x)/2 ⇒ 8 = 2 + x ⇒ x = 6
Thus, the missing coordinate is x = 6.
4. Real-life Connection
Midpoints are useful in geometry, design, construction, and computer graphics. For example, to find the centre of a bridge span or a midpoint in digital mapping.
Classwork / Practice Exercises
- Find the midpoint of the line joining the points (–2, 5) and (4, 9).
- Find the midpoint of the line joining (3, 7) and (9, –1).
- If the midpoint of AB is (2, 3) and A(4, 5), find the coordinates of B.
- The midpoint of PQ is (1, –2). If P(5, 0), find the coordinates of Q.
- The midpoint of a line joining (x, 4) and (10, –2) is (6, 1). Find x.
Assignment
- If A(–3, 2) and B(5, k) have a midpoint (1, 5), find the value of k.
- The midpoint of CD is (2, –1). If C(–4, 3), find the coordinates of D.
- A line has endpoints at (–2, –6) and (6, 4). Find the midpoint and verify your answer graphically.