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IB MYP 4-5 Maths-Gradients of perpendicular lines- Study Notes

IB MYP 4-5 Maths- Gradients of perpendicular lines- Study Notes - New Syllabus

IB MYP 4-5 Maths- Gradients of perpendicular lines – Study Notes

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  • Gradients of perpendicular lines

IB MYP 4-5 Maths- Gradients of perpendicular lines – Study Notes – All topics

Perpendicular Lines

Perpendicular Lines

Two lines are perpendicular if they intersect at a right angle (90°). The product of their gradients is:

\( m_1 \times m_2 = -1 \)

This means if one line has a gradient \( m_1 \), then the perpendicular line has a gradient:

\( m_2 = -\dfrac{1}{m_1} \)

Key Points:

  • Condition: \( m_1 \cdot m_2 = -1 \).
  • If the first line is horizontal (\( m = 0 \)), the perpendicular line is vertical (undefined slope).
  • Perpendicular lines form a right angle at their intersection point.

Question:

Find the equation of a line that is perpendicular to \( y = 2x + 3 \) and passes through the point (4, 1).

▶️ Answer/Explanation

Step 1: Gradient of given line: \( m_1 = 2 \).

Step 2: Gradient of perpendicular line: \( m_2 = -\dfrac{1}{m_1} = -\dfrac{1}{2} \).

Step 3: Equation using point-slope form:

\( y – 1 = -\dfrac{1}{2}(x – 4) \)

\( y – 1 = -\dfrac{1}{2}x + 2 \)

\( y = -\dfrac{1}{2}x + 3 \)

Question:

A line passes through (0, 5) and is perpendicular to the line joining (2, 3) and (6, 7). Find its equation.

▶️ Answer/Explanation

Step 1: Find gradient of line through (2, 3) and (6, 7):

\( m_1 = \dfrac{7 – 3}{6 – 2} = \dfrac{4}{4} = 1 \).

Step 2: Gradient of required line:

\( m_2 = -\dfrac{1}{m_1} = -1 \).

Step 3: Use point (0, 5):

\( y – 5 = -1(x – 0) \)

\( y = -x + 5 \)

Question :

Find the equation of the perpendicular bisector of the segment joining A(2, -1) and B(6, 3).

▶️ Answer/Explanation

Step 1: Find midpoint of AB:

\( M = \left( \dfrac{2+6}{2}, \dfrac{-1+3}{2} \right) = (4, 1) \).

Step 2: Gradient of AB:

\( m_1 = \dfrac{3 – (-1)}{6 – 2} = \dfrac{4}{4} = 1 \).

Step 3: Gradient of perpendicular bisector:

\( m_2 = -\dfrac{1}{m_1} = -1 \).

Step 4: Equation using point (4, 1):

\( y – 1 = -1(x – 4) \)

\( y – 1 = -x + 4 \)

\( y = -x + 5 \)

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