Parallel lines

1. Overview

Parallel lines are straight lines that never intersect. The key property of parallel lines is that they have the same gradient. This revision note covers how to identify parallel lines from their equations, and how to find the equation of a line parallel to a given line that passes through a specific point. This is a fundamental concept in coordinate geometry.

Key Definitions

• Parallel: Two or more lines that are always the same distance apart and never intersect.
• Gradient ($m$): A measure of the steepness of a line, calculated as $\frac{\text{change in } y}{\text{change in } x}$.
• $y$-intercept ($c$): The point where a line crosses the $y$-axis (where $x = 0$).
• Equation of a straight line: Usually written in the form $y = mx + c$.

Core Content

The Fundamental Rule of Parallel Lines

The most important rule to remember is: Parallel lines have the same gradient.

If two lines $L_1$ and $L_2$ are parallel, then:

$\qquad m_1 = m_2$

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Finding the Equation of a Parallel Line

To find the equation of a line parallel to a given line that passes through a specific point $(x, y)$, follow these steps:

1. Identify the gradient ($m$) of the given line. (If the line is not in the form $y = mx + c$, you must rearrange it first).
2. Use the same gradient for your new line.
3. Substitute the coordinates of the given point $(x, y)$ and the gradient $(m)$ into the general formula $y = mx + c$ to find the new $y$-intercept ($c$).
4. Rewrite the final equation using the $m$ and the new $c$.

Worked example 1 — Finding the equation

Find the equation of the line parallel to $y = 4x - 7$ that passes through the point $(2, 11)$.

• Step 1: Identify the gradient. The given line is $y = 4x - 7$. The gradient $m = 4$.
• Step 2: Use the same gradient for the new line. Parallel lines have equal gradients, so the new gradient is also $m = 4$.
• Step 3: Substitute the point $(2, 11)$ into $y = mx + c$. $11 = 4(2) + c$ $11 = 8 + c$
• Step 4: Solve for $c$. $c = 11 - 8$ $c = 3$
• Step 5: Write the final equation. $\qquad \boxed{y = 4x + 3}$

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Identifying Parallel Lines from Complex Equations

Sometimes the gradient is not immediately obvious. You must rearrange the equation into the form $y = mx + c$ by making $y$ the subject.

Worked example 2 — Rearranging first

Show that the line $2y - 6x = 10$ is parallel to the line $y = 3x + 1$.

• Step 1: Rearrange the first equation to $y = mx + c$. $2y - 6x = 10$ $2y = 6x + 10$ (Add $6x$ to both sides) $y = 3x + 5$ (Divide every term by 2)
• Step 2: Compare gradients. The gradient of the first line is $m = 3$. The gradient of the second line ($y = 3x + 1$) is $m = 3$.
• Conclusion: Because both lines have a gradient of 3, they are parallel.

Worked example 3 — Parallel line through a fractional coordinate

Find the equation of the line that is parallel to $y = -\frac{1}{2}x + 5$ and passes through the point $(\frac{2}{3}, 4)$.

• Step 1: Identify the gradient of the given line. The given line is $y = -\frac{1}{2}x + 5$. Therefore, the gradient $m = -\frac{1}{2}$.
• Step 2: Use the same gradient for the parallel line. Since the lines are parallel, the gradient of the new line is also $m = -\frac{1}{2}$.
• Step 3: Substitute the point $(\frac{2}{3}, 4)$ and the gradient $m = -\frac{1}{2}$ into $y = mx + c$. $4 = (-\frac{1}{2})(\frac{2}{3}) + c$ $4 = -\frac{1}{3} + c$
• Step 4: Solve for $c$. $c = 4 + \frac{1}{3}$ (Add $\frac{1}{3}$ to both sides) $c = \frac{12}{3} + \frac{1}{3}$ $c = \frac{13}{3}$
• Step 5: Write the final equation. $\qquad \boxed{y = -\frac{1}{2}x + \frac{13}{3}}$

Extended Content (Extended Only)

While the core content focuses on finding the equation of a parallel line given a point and another line's equation, the Extended curriculum expects a deeper understanding of how parallel lines interact with other geometric concepts. For instance, you might be asked to find the equation of a line parallel to a given line that forms a specific shape (like a parallelogram) with other lines. This requires combining your knowledge of parallel lines with area calculations and geometric properties.

Worked example 4 — Parallel line forming a parallelogram

Lines $L_1$ and $L_2$ have equations $y = 2x + 1$ and $y = 2x - 3$ respectively. Line $L_3$ passes through the point $(0, 5)$ and intersects $L_1$ at point $A$ and $L_2$ at point $B$. Given that $L_3$ is parallel to the x-axis, find the equation of a line $L_4$ that is parallel to $L_1$ and forms a parallelogram $ABCD$ with $L_1$, $L_2$, and $L_3$, where $CD$ lies on $L_3$.

• Step 1: Find the coordinates of points A and B. Since $L_3$ is parallel to the x-axis and passes through $(0, 5)$, its equation is $y = 5$. To find point A (intersection of $L_1$ and $L_3$): $5 = 2x + 1$ $4 = 2x$ $x = 2$. So, $A = (2, 5)$. To find point B (intersection of $L_2$ and $L_3$): $5 = 2x - 3$ $8 = 2x$ $x = 4$. So, $B = (4, 5)$.

• Step 2: Determine the length of AB. $AB = 4 - 2 = 2$ units.

• Step 3: Understand the properties of a parallelogram. In a parallelogram, opposite sides are equal in length and parallel. Since $ABCD$ is a parallelogram and $L_4$ is parallel to $L_1$ and $L_2$, the length of $CD$ must also be 2 units. Also, $CD$ lies on $L_3$ (y=5).

• Step 4: Find the x-coordinate of point C. Let $C = (x, 5)$. Since $CD = 2$ and $D$ lies on $L_4$, we need to find a point on $L_2$ such that the horizontal distance to $C$ is 2. Since $B = (4, 5)$, we can assume that $C$ is to the left of $B$. $x = 4 - 2 = 2$. So, $C = (2, 5)$.

• Step 5: Find the x-coordinate of point D. Since $A = (2, 5)$ and $CD = 2$, we can assume that $D$ is to the left of $A$. $x = 2 - 2 = 0$. So, $D = (0, 5)$.

• Step 6: Find the y-intercept of $L_4$. Since $L_4$ passes through $D = (0, 5)$, the y-intercept of $L_4$ is 5.

• Step 7: Write the equation of $L_4$. Since $L_4$ is parallel to $L_1$, it has the same gradient, which is 2. Therefore, the equation of $L_4$ is: $\qquad \boxed{y = 2x + 5}$

Key Equations

$\qquad \textbf{The General Equation: } \mathbf{y = mx + c}$

• $y$: $y$-coordinate
• $x$: $x$-coordinate
• $m$: Gradient (slope)
• $c$: $y$-intercept

Note: This formula is not provided on the IGCSE formula sheet. You must memorise it.

$\qquad \textbf{Parallel Condition: } \mathbf{m_1 = m_2}$

• The gradient of the first line is equal to the gradient of the second line.

Note: This formula is not provided on the IGCSE formula sheet. You must memorise it.

Common Mistakes to Avoid

• ❌ Confusing Gradient and y-intercept: Using the $y$-intercept ($c$) of the given line as the gradient ($m$) for the new parallel line.
  • ✓ Right: Always identify the coefficient of $x$ as the gradient ($m$) when the equation is in the form $y = mx + c$. The $y$-intercept is only used to find the new $y$-intercept of the parallel line.
• ❌ Incomplete Rearranging: When rearranging an equation to the form $y = mx + c$, forgetting to divide every term by the coefficient of $y$.
  • Example: For $3y = 6x + 9$, incorrectly stating $m = 6$.
  • ✓ Right: Divide the entire equation by 3 to get $y = 2x + 3$, so $m = 2$.
• ❌ Altering the Gradient for Parallel Lines: Changing the sign or inverting the gradient when finding a parallel line. This is only for perpendicular lines.
  • ✓ Right: The gradient of a parallel line is exactly the same as the original line.

Exam Tips

• Command Words: Look for "Find the equation of a line parallel to..." or "Show that these lines are parallel...". If the question says "Show that," you must explicitly state the gradients of both lines and a concluding statement at the end of your working.
• Calculator Use: While these are often non-calculator questions, you can use a calculator to solve for $c$ if the coordinates are fractions or decimals, or to check your rearrangement of equations.
• Show Your Working: Marks are awarded for:
  1. Correctly identifying (or calculating) the gradient.
  2. Substituting the given point into $y = mx + c$.
  3. Correctly calculating the new $y$-intercept ($c$).
  4. Writing the final equation in the form $y = mx + c$.
• Hidden Gradients: If you are given two points instead of an equation, first find the gradient using $m = \frac{y_2 - y_1}{x_2 - x_1}$, then apply the parallel rule.