Equations of Lines (HSC SSCE Mathematics Advanced): Revision Notes

Equations of Lines

Introduction

Finding the equation of a straight line is a fundamental skill in coordinate geometry. The equation describes all points that lie on that line. In this topic, you will learn several methods to determine line equations depending on the information given.

Point-gradient form

When you know a line's gradient and one point it passes through, you can use the point-gradient formula to find the complete equation of the line.

Point-gradient form: The equation of a line with a known gradient and point can be written as $y - y_1 = m(x - x_1)$.

The formula is:

$y - y_1 = m(x - x_1)$

Where:

• $m$ is the gradient of the line
• $(x_1, y_1)$ is a point on the line

To use this formula, substitute the given gradient and coordinates into the equation, then rearrange to gradient-intercept form $y = mx + c$ if required.

Line through two points

When you have two points on a line but no gradient given directly, you must first calculate the gradient before finding the equation.

To find the equation of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$:

Step 1: Calculate the gradient using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where:

• $m$ is the gradient of the line
• $(x_1, y_1)$ are the coordinates of the first point
• $(x_2, y_2)$ are the coordinates of the second point

Step 2: Use the point-gradient formula $y - y_1 = m(x - x_1)$ with one of the points and the calculated gradient, then rearrange to $y = mx + c$.

Parallel lines

Parallel lines are lines that run in the same direction and never meet, no matter how far they are extended. Understanding parallel lines is essential for solving many geometric problems.

Two lines with equations $y = m_1x + c_1$ and $y = m_2x + c_2$ are parallel if their gradients are equal:

$m_1 = m_2$

If $c_1 = c_2$ as well, the two lines are identical rather than distinct parallel lines.

Finding parallel line equations

To determine the equation of a straight line parallel to a given line and passing through a specific point $(x_1, y_1)$:

1. Identify the gradient $m$ from the given line
2. Use the point-gradient form $y - y_1 = m(x - x_1)$ with the given point $(x_1, y_1)$ and the same gradient $m$
3. Rearrange into gradient-intercept form $y = mx + c$ or general form $ax + by + c = 0$ as required

For example, both lines shown have gradient $m = 2$ but different y-intercepts ($c = -1$ and $c = 3$), so they are parallel.

Perpendicular lines

Perpendicular lines intersect at right angles, forming a 90-degree corner where they meet. This relationship creates a special connection between their gradients.

Perpendicular lines: Two straight lines that intersect at a 90° angle (a right angle).

If the gradients of two lines are $m_1$ and $m_2$, they are perpendicular if:

$m_1 \times m_2 = -1$

Finding perpendicular line equations

To determine the equation of a line perpendicular to a given line through a specific point:

1. Determine the gradient $m$ of the given line
2. Calculate the gradient of the perpendicular line using $m_2 = -\frac{1}{m}$
3. Use the point-gradient form with the given point $(x_1, y_1)$: $y - y_1 = m_2(x - x_1)$
4. Rearrange into gradient-intercept or general form as required

For example, on this graph, the gradients of the two straight lines are $m_1 = 2$ and $m_2 = -\frac{1}{2}$. Their product is $2 \times \left(-\frac{1}{2}\right) = -1$, confirming the two lines are perpendicular.

Applications of parallel and perpendicular lines

Parallel and perpendicular lines appear frequently in real-world contexts. Engineers use them when designing roads and bridges, architects apply them when planning building structures, and physicists use them to model scenarios like parallel roads, perpendicular walls, or reflected light paths in optics. These applications often involve specific constraints, such as fixed points or particular angles.

Problem-solving approach

To solve problems involving parallel or perpendicular lines:

1. Identify the given line's equation and determine its gradient $m$
2. For parallel lines, use the same gradient; for perpendicular lines, use the negative reciprocal $-\frac{1}{m}$
3. Apply the point-gradient form with any given points or constraints, and consider the context (e.g., horizontal/vertical lines)
4. Verify the solution geometrically (checking angles or intercepts) or algebraically

Remember!