IB MYP 4-5 Maths- Gradient of parallel lines - Study Notes - New Syllabus
IB MYP 4-5 Maths- Gradient of parallel lines – Study Notes
Standard
- Gradient of parallel lines
IB MYP 4-5 Maths- Gradient of parallel lines – Study Notes – All topics
Parallel Lines
Parallel Lines
Two lines are parallel if they have the same gradient (slope) but different y-intercepts.
Condition for Parallelism:
\( \text{If } m_1 = m_2, \text{ then the lines are parallel.} \)
Key Points:
- Parallel lines never intersect.
- They have equal gradients, but different y-intercepts.
- If two lines have the same equation, they are coincident (infinite solutions).
Example:
Find the equation of the line parallel to \( y = 2x + 3 \) and passing through (4, 5).
▶️ Answer/Explanation
The given line has gradient \( m = 2 \).
Parallel line also has \( m = 2 \).
Use point-slope form: \( y – y_1 = m(x – x_1) \)
\( y – 5 = 2(x – 4) \)
\( y – 5 = 2x – 8 \)
\( y = 2x – 3 \)
Example :
Determine whether the lines \( y = 3x + 7 \) and \( y = 3x – 4 \) are parallel.
▶️ Answer/Explanation
The first line has gradient \( m_1 = 3 \).
The second line has gradient \( m_2 = 3 \).
Since \( m_1 = m_2 \) and intercepts are different, the lines are parallel.
Example:
Find the equation of the line passing through the point (4, 3) and parallel to the line \( y = 2x – 5 \).
▶️ Answer/Explanation
Step 1: The given line has gradient \( m = 2 \). For a parallel line, gradient remains the same.
Step 2: Use point-slope form with point (4, 3):
\( y – 3 = 2(x – 4) \)
\( y – 3 = 2x – 8 \)
\( y = 2x – 5 \)
Example:
The line L1 has equation \( y = -\dfrac{1}{2}x + 4 \). Find the equation of the line parallel to L1 that passes through the intersection of the lines \( y = x + 1 \) and \( y = 3x – 5 \).
▶️ Answer/Explanation
Step 1: Find the intersection point of \( y = x + 1 \) and \( y = 3x – 5 \):
\( x + 1 = 3x – 5 \Rightarrow 6 = 2x \Rightarrow x = 3 \).
Substitute in \( y = x + 1 \): \( y = 3 + 1 = 4 \).
Intersection point: (3, 4).
Step 2: Gradient of L1 is \( m = -\dfrac{1}{2} \).
Step 3: Equation of required line through (3, 4):
\( y – 4 = -\dfrac{1}{2}(x – 3) \)
\( y – 4 = -\dfrac{1}{2}x + \dfrac{3}{2} \)
\( y = -\dfrac{1}{2}x + \dfrac{3}{2} + 4 = -\dfrac{1}{2}x + \dfrac{11}{2} \)