IB Mathematics AA SL equation of a straight line SL Study Notes
IB Mathematics AA SL equation of a straight line SL Study Notes Offer a clear explanation of equation of a straight line , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic equation of a straight line based on IB math syllabus
Different Forms of the Equation of a Straight Line
Different Forms of the Equation of a Straight Line
In coordinate geometry, the equation of a straight line can be written in various forms depending on the information given. The most common forms are:
- 1. Gradient-Intercept Form:
\( y = mx + c \)
where m is the gradient (slope) of the line and c is the y-intercept (the point where the line crosses the y-axis).
- 1. Gradient-Intercept Form:
- 2. Standard Form:
\( ax + by + d = 0 \)
where a, b, and d are constants. This is the standard linear form in two variables.
- 2. Standard Form:
- 3. Point-Gradient (Point-Slope) Form:
\( y – y_1 = m(x – x_1) \)
where (x₁, y₁) is a point on the line, and m is the gradient.
Each form is useful in different contexts. For example, use the gradient-intercept form when you know the slope and intercept, and use the point-gradient form when you know a point and the slope.
Example :
Find the equation of the line with gradient \( m = 2 \) and y-intercept \( c = -3 \).
▶️Answer/Explanation
We use the form \( y = mx + c \)
\( y = 2x – 3 \)
Example :
Write the equation \( y = 3x + 4 \) in general form.
▶️Answer/Explanation
Rearrange to move all terms to one side:
\( y – 3x – 4 = 0 \) or \( -3x + y – 4 = 0 \)
Example :
Find the equation of the line passing through the point \( (2, 5) \) with a gradient of \( 4 \).
▶️Answer/Explanation
Use the formula \( y – y_1 = m(x – x_1) \)
\( y – 5 = 4(x – 2) \)
Expand: \( y = 4x – 8 + 5 = 4x – 3 \)
Lines with Gradients \( m_1 \) and \( m_2 \)
Lines with Gradients \( m_1 \) and \( m_2 \)
The gradient (or slope) of a straight line indicates its steepness. The relationship between the gradients of two lines reveals whether they are parallel or perpendicular:
1. Parallel Lines:
Two lines are parallel if they have the same gradient.
Condition:\( m_1 = m_2 \)
2. Perpendicular Lines:
Two lines are perpendicular if the product of their gradients is −1.
Condition:\( m_1 \cdot m_2 = -1 \)
This means the gradients are negative reciprocals of each other.
Example :
Determine if the lines \( y = 2x + 3 \) and \( y = 2x – 5 \) are parallel.
▶️Answer/Explanation
Both lines are in the form \( y = mx + c \) and have gradient \( m = 2 \).
Since the gradients are equal, the lines are parallel.
Example :
Are the lines \( y = 3x + 1 \) and \( y = -\frac{1}{3}x + 4 \) perpendicular?
▶️Answer/Explanation
Gradient of the first line is \( m_1 = 3 \)
Gradient of the second line is \( m_2 = -\frac{1}{3} \)
Check: \( m_1 \cdot m_2 = 3 \cdot -\frac{1}{3} = -1 \)
Since the product is −1, the lines are perpendicular.
Gradient of Inclines: Real-Life Applications
Gradient (Slope) Formula: The gradient between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\( m = \frac{y_2 – y_1}{x_2 – x_1} \)
Interpretation:
- If \( m > 0 \): uphill incline
- If \( m < 0 \): downhill incline
- If \( m = 0 \): flat surface
Applications: Gradient is commonly used in engineering and geography for designing roads, bridges, and railway tracks.
Example:
A road climbs from a height of 200 m to 450 m over a horizontal distance of 1.5 km. Find the gradient of the road.
▶️Answer/Explanation
Convert horizontal distance to meters:
\( \text{Run} = 1.5 \, \text{km} = 1500 \, \text{m} \)
\( \text{Rise} = 450 – 200 = 250 \, \text{m} \)
Gradient:
\( m = \frac{\text{rise}}{\text{run}} = \frac{250}{1500} = \frac{1}{6} \approx 0.167 \)
Interpretation: The road rises about 1 meter for every 6 meters of horizontal distance, indicating a mild incline.