In this section, we explore how to derive the equation of a straight line. Understanding these methods is crucial for solving problems in coordinate geometry. We will cover two primary methods: using two points (via the two-point formula) and using a single point with a known gradient (point-slope form).

**The General Form of a Line Equation**

- A line equation is typically expressed as $y = mx + c$.
- Here, $m$ represents the gradient of the line.
- $c$ is the y-intercept, the point where the line crosses the y-axis.

**Method 1: Using a Point and the Gradient (Point-Slope Form)**

When we know a point $A(x_1, y_1)$ on the line and its gradient $m$, we can express the line's equation as: $y - y_1 = m(x - x_1)$

**Example 1**

Find the equation of the straight line with a gradient of 3 that passes through the point (1, 6).

**Solution: **

Let $A(1, 6)$ and $m = 3$.

The line equation becomes:

$y - 6 = 3(x - 1)$

Expanding this, we get:

$y = 3x - 3 + 6$

Therefore, the line equation is:

$\therefore y = 3x + 3$

**Method 2: Using Two Points (Two-Point Formula)**

When two points, say $(x_1, y_1)$ and $(x_2, y_2)$, are given, we first find the gradient $m$using: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Then, we use this gradient with one of the points in the point-slope form to find the equation.

**Example 2**

Determine the equation of the line passing through the points (-5, 3) and (-4, 1).

**Solution:**

First, calculate the gradient $m$:

$m = \frac{1 - 3}{-4 - (-5)} = \frac{-2}{1} = -2$

The general form of the line is $y = -2x + c$.

To find $c$, substitute one of the points, say (-4, 1):

$1 = -2(-4) + c$

$1 = 8 + c$

$c = -7$

Thus, the equation of the line is:

$\therefore y = -2x - 7$

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.