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
Equation of a Straight Line
The equation of a straight line describes a linear relationship between the variables \(x\) and \(y\). The most common forms of this equation are the slope-intercept form and the point-slope form.
1. Slope-Intercept Form
The slope-intercept form of a line’s equation is given by:
\( y = mx + c \)
- \(m\): The slope of the line, representing the rate of change of \(y\) with respect to \(x\).
- \(c\): The y-intercept, where the line crosses the y-axis (\(x = 0\)).
Example 1: Finding the Equation
Find the equation of a line with a slope \(m = 2\) and a y-intercept \(c = -3\).
Using the formula \( y = mx + c \):
\( y = 2x – 3 \)
This is the equation of the line with slope 2 and y-intercept -3.
2. Point-Slope Form
The point-slope form of a line’s equation is given by:
\( y – y_1 = m(x – x_1) \)
- \(m\): The slope of the line.
- \((x_1, y_1)\): A point on the line.
Example 2: Finding the Equation Using Point-Slope Form
Find the equation of a line passing through the point \((3, 4)\) with a slope \(m = -1\).
Using the formula \( y – y_1 = m(x – x_1) \):
\( y – 4 = -1(x – 3) \)
Simplifying this equation:
\( y – 4 = -x + 3 \)
\( y = -x + 7 \)
This is the equation of the line in slope-intercept form.
3. General Form
The general form of a line’s equation is given by:
\( Ax + By + C = 0 \)
- \(A\), \(B\), \(C\): Constants, where \(A\) and \(B\) are not both zero.
Example 3: Converting to General Form
Convert the equation \( y = 2x + 5 \) to general form.
Rearranging terms:
\( 2x – y + 5 = 0 \)
So, the general form of the equation is \( 2x – y + 5 = 0 \).
4. Finding Slope from Two Points
If two points on a line are given as \((x_1, y_1)\) and \((x_2, y_2)\), the slope \(m\) can be calculated using:
\( m = \frac{y_2 – y_1}{x_2 – x_1} \)
Example 4: Calculating Slope
Find the slope of the line passing through the points \((1, 2)\) and \((4, 6)\).
Using the formula \( m = \frac{y_2 – y_1}{x_2 – x_1} \):
\( m = \frac{6 – 2}{4 – 1} = \frac{4}{3} \)
So, the slope of the line is \(\frac{4}{3}\).
Key Takeaways
- Slope-Intercept Form: Best for quickly identifying the slope and y-intercept of the line.
- Point-Slope Form: Useful when you know a point on the line and the slope.
- General Form: A standard representation that is useful in various mathematical applications.
- Finding Slope: Calculating the slope using two points helps in forming the line equation.
Example Problems: Equation of a Straight Line
Example 1: Finding the Slope-Intercept Form
Problem: Find the equation of a line with a slope of \(m = 3\) that passes through the point \((0, -2)\).
Solution:
Since the line passes through the y-intercept \((0, -2)\), we have \(c = -2\). Using the slope-intercept form \(y = mx + c\):
\(y = 3x – 2\)
Answer: The equation of the line is \(y = 3x – 2\).
Example 2: Using Point-Slope Form
Problem: Write the equation of a line passing through the point \((2, 5)\) with a slope \(m = -4\).
Solution:
Using the point-slope form \(y – y_1 = m(x – x_1)\):
\(y – 5 = -4(x – 2)\)
Expanding this equation:
\(y – 5 = -4x + 8\)
\(y = -4x + 13\)
Answer: The equation of the line is \(y = -4x + 13\).
Example 3: Finding the Slope from Two Points
Problem: Determine the slope of a line passing through the points \((1, 3)\) and \((4, 11)\), and find its equation.
Solution:
First, calculate the slope using the formula \(m = \frac{y_2 – y_1}{x_2 – x_1}\):
\(m = \frac{11 – 3}{4 – 1} = \frac{8}{3}\)
Using the slope-intercept form \(y – y_1 = m(x – x_1)\) with point \((1, 3)\):
\(y – 3 = \frac{8}{3}(x – 1)\)
Expanding this equation:
\(y – 3 = \frac{8}{3}x – \frac{8}{3}\)
\(y = \frac{8}{3}x – \frac{8}{3} + 3\)
\(y = \frac{8}{3}x + \frac{1}{3}\)
Answer: The equation of the line is \(y = \frac{8}{3}x + \frac{1}{3}\).
Example 4: Converting Slope-Intercept to General Form
Problem: Convert the equation \(y = 5x – 7\) into the general form.
Solution:
Rearrange the terms to express it in the form \(Ax + By + C = 0\):
\(5x – y – 7 = 0\)
Answer: The general form of the equation is \(5x – y – 7 = 0\).
Example 5: Determining the Equation of a Horizontal Line
Problem: Find the equation of a line passing through the point \((6, -4)\) that is parallel to the x-axis.
Solution:
A line parallel to the x-axis has a slope \(m = 0\). The equation of a horizontal line is \(y = k\), where \(k\) is the y-coordinate of any point on the line.
Thus, the equation is:
\(y = -4\)
Answer: The equation of the horizontal line is \(y = -4\).
IB Mathematics AA SL Equation of a Straight Line Exam Style Worked Out Questions
Question
Point P has coordinates (-3 , 2), and point Q has coordinates (15, -8). Point M is the midpoint of [PQ].
(a) Find the coordinates of M.
Line L is perpendicular to [PQ] and passes through M.
(b) Find the gradient of L.
(c) Hence, write down the equation of L.
▶️Answer/Explanation
Answer:
(a) M (6, -3)
(b) gradient of [PQ] = -\(\frac{5}{9}\)
gradient of L = \(\frac{5}{9}\)
(c) y + 3 = \(\frac{9}{5}\)(x – 6) OR y = \(\frac{9}{5}\) x – \(\frac{69}{5}\) (or equivalent)
Question
Consider the points A (-2 , 20) , B (4 , 6) and C (-14 , 12) . The line L passes through the point A and is perpendicular to [BC] .
(a) Find the equation of L . [3]
The line L passes through the point (k , 2) .
(b) Find the value of k .
▶️Answer/Explanation
b.
substituting (k , 2) into their L
2 − 20 = 3(k + 2)
OR
2 = 3k + 26
k=-8