Perpendicular lines

1. Overview

Perpendicular lines are lines that intersect at a right angle ($90^{\circ}$). The key concept for IGCSE Maths is understanding the relationship between their gradients: the gradient of a line perpendicular to another is its negative reciprocal. This allows you to determine the equation of a perpendicular line, given the equation of the original line and a point it passes through. This topic is crucial for solving coordinate geometry problems.

Key Definitions

• Perpendicular: Two lines that intersect at exactly $90^{\circ}$.
• Gradient ($m$): A measure of the steepness of a line, defined as $\frac{\text{change in } y}{\text{change in } x}$. Also commonly referred to as slope.
• Negative Reciprocal: The value obtained by flipping a fraction and changing its sign. For a gradient $m$, the negative reciprocal is $-\frac{1}{m}$.
• Product: The result of multiplying two numbers together.

Core Content

There are no specific Core-only objectives for this sub-topic. All learning objectives regarding the calculation of perpendicular gradients are part of the Supplement (Extended) curriculum.

Extended Content (Extended Only)

The fundamental rule for perpendicular lines with gradients $m_1$ and $m_2$ is:

$\qquad \boxed{m_1 \times m_2 = -1}$

This means that the gradient of a perpendicular line is the negative reciprocal of the original line's gradient. Understanding and applying this relationship is essential for solving problems involving perpendicular lines.

Method: Finding the Perpendicular Gradient

1. Identify the gradient of the original line ($m_1$).
2. If the gradient is a whole number like $3$, think of it as $\frac{3}{1}$.
3. Flip the fraction and change the sign to find $m_2$.

Numerical Example:

• If $m_1 = \frac{2}{3}$, then $m_2 = -\frac{3}{2}$
• If $m_1 = -4$, then $m_2 = +\frac{1}{4}$
• If $m_1 = 1$, then $m_2 = -1$

---

Worked Example 1 — Finding the Equation of a Perpendicular Line

Question: Find the equation of the line perpendicular to $y = 2x + 5$ that passes through the point $(4, 7)$.

Step 1: Identify the gradient of the given line ($m_1$). The equation is in the form $y = mx + c$. $\qquad y = 2x + 5$ Therefore, $\qquad m_1 = 2$

Step 2: Calculate the perpendicular gradient ($m_2$). Using $m_2 = -\frac{1}{m_1}$: $\qquad m_2 = -\frac{1}{2}$

Step 3: Use the point $(4, 7)$ and $m_2$ to find the new equation. Substitute $x = 4$, $y = 7$, and $m = -\frac{1}{2}$ into $y = mx + c$: $\qquad 7 = (-\frac{1}{2})(4) + c$ $\qquad 7 = -2 + c$ Add 2 to both sides: $\qquad 7 + 2 = -2 + 2 + c$ $\qquad 9 = c$

Step 4: Write the final equation. $\qquad \boxed{y = -\frac{1}{2}x + 9}$

---

Worked Example 2 — Perpendicular lines from a general form equation

Question: Line $L_1$ has the equation $3x + 4y = 12$. Find the gradient of a line perpendicular to $L_1$.

Step 1: Rearrange $L_1$ into $y = mx + c$ form to find the gradient. $\qquad 3x + 4y = 12$ Subtract $3x$ from both sides: $\qquad 4y = -3x + 12$ Divide every term by 4: $\qquad y = -\frac{3}{4}x + 3$ So, $\qquad m_1 = -\frac{3}{4}$

Step 2: Find the negative reciprocal. Flip the fraction and change the sign: $\qquad m_2 = \frac{4}{3}$

Answer: The perpendicular gradient is $\boxed{\frac{4}{3}}$ (or $1.33$ to 3sf).

---

Worked Example 3 — Finding the equation given two points on the perpendicular line

Question: Line $L$ passes through the points $(1, 5)$ and $(4, -1)$. Find the equation of the line perpendicular to $L$ that passes through the point $(2, 3)$.

Step 1: Calculate the gradient of line $L$ ($m_1$). Using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$: $\qquad m_1 = \frac{-1 - 5}{4 - 1}$ $\qquad m_1 = \frac{-6}{3}$ $\qquad m_1 = -2$

Step 2: Calculate the perpendicular gradient ($m_2$). Using $m_2 = -\frac{1}{m_1}$: $\qquad m_2 = -\frac{1}{-2}$ $\qquad m_2 = \frac{1}{2}$

Step 3: Use the point $(2, 3)$ and $m_2$ to find the new equation. Substitute $x = 2$, $y = 3$, and $m = \frac{1}{2}$ into $y = mx + c$: $\qquad 3 = (\frac{1}{2})(2) + c$ $\qquad 3 = 1 + c$ Subtract 1 from both sides: $\qquad 3 - 1 = 1 - 1 + c$ $\qquad 2 = c$

Step 4: Write the final equation. $\qquad \boxed{y = \frac{1}{2}x + 2}$

---

Worked Example 4 — Showing that two lines are perpendicular

Question: Line $L_1$ has equation $y = 3x - 2$. Line $L_2$ passes through points $(0, 4)$ and $(3, 3)$. Show that $L_1$ and $L_2$ are perpendicular.

Step 1: Find the gradient of $L_1$. The equation is in the form $y = mx + c$, so the gradient is simply the coefficient of $x$. $\qquad m_1 = 3$

Step 2: Find the gradient of $L_2$. Using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$: $\qquad m_2 = \frac{3 - 4}{3 - 0}$ $\qquad m_2 = \frac{-1}{3}$ $\qquad m_2 = -\frac{1}{3}$

Step 3: Show that the product of the gradients is -1. $\qquad m_1 \times m_2 = 3 \times (-\frac{1}{3})$ $\qquad m_1 \times m_2 = -1$

Answer: Since the product of the gradients is -1, the lines $L_1$ and $L_2$ are perpendicular.

Key Equations

• Perpendicular Gradient Rule: $\qquad \boxed{m_1 \times m_2 = -1}$

  • $m_1$: Gradient of the first line
  • $m_2$: Gradient of the second line

• Gradient through two points: $\qquad \boxed{m = \frac{y_2 - y_1}{x_2 - x_1}}$

• Equation of a straight line: $\qquad \boxed{y = mx + c}$

  • $m$: Gradient
  • $c$: y-intercept

Note: These formulas are not provided on the IGCSE formula sheet; they must be memorized.

Common Mistakes to Avoid

• ❌ Wrong: Thinking perpendicular lines have the same gradient.
  • ✓ Right: Parallel lines have the same gradient ($m_1 = m_2$); perpendicular lines have negative reciprocal gradients ($m_1 \times m_2 = -1$).
• ❌ Wrong: Only flipping the fraction but forgetting to change the sign when finding the negative reciprocal.
  • ✓ Right: If $m = \frac{1}{2}$, the perpendicular gradient is $-2$, not $+\frac{2}{1}$. The sign must change.
• ❌ Wrong: Forgetting to rearrange equations into $y = mx + c$ form before identifying the gradient.
  • ✓ Right: In the equation $2y = 6x + 4$, the gradient is NOT 6. You must divide by 2 first to get $y = 3x + 2$, so $m = 3$.
• ❌ Wrong: Confusing the $x$ and $y$ values when calculating the gradient from two points.
  • ✓ Right: Always use $m = \frac{y_2 - y_1}{x_2 - x_1}$, ensuring that the $y$ values are in the numerator and the corresponding $x$ values are in the denominator.

Exam Tips

• Show your working: Even if you can find the negative reciprocal in your head, write down "$-1 \div (\text{your gradient})$" to secure method marks if you make a calculation error.
• Command Words: If a question asks you to "Show that" two lines are perpendicular, calculate both gradients separately and then show that their product is $-1$. Do not just state it; prove it mathematically.
• Calculator use: When finding the negative reciprocal of a decimal, use the $x^{-1}$ key on your calculator and then change the sign. For example, if $m = 0.8$, type 0.8, press x⁻¹, then change the sign to get $-1.25$.
• Real-world context: These problems often appear in geometry questions involving tangents to circles or finding the shortest distance from a point to a line. Remember: the shortest distance is always the perpendicular distance.
• Double-check your arithmetic: A simple arithmetic error when calculating the gradient or substituting values into $y = mx + c$ can lead to an incorrect answer. Take a moment to review your calculations.