Coordinate Geometry of Straight Lines for the TMUA

The Gradient and Linear Equations

Every straight line can be defined by its gradient $m$ and its position on the $(x, y)$ plane. The most familiar form is $y = mx + c$, where $m$ represents the gradient and $c$ is the $y$-intercept (the value of $y$ where the line crosses the $y$-axis).

Interpreting the Gradient

The gradient $m$ measures the steepness of a line. If $m$ is positive, the line slopes upwards from bottom left to top right. If $m$ is negative, it slopes downwards from top left to bottom right. We can think of steepness as a rate of change: for every 1 unit move horizontally (in the $x$ direction), we must move $m$ units vertically (in the $y$ direction) to remain on the line.

In the diagram above, the gradient is 2. For every 1 unit of $x$ increased, $y$ increases by 2.

In this second diagram, the gradient is -3. For every 1 unit of $x$ increased, $y$ decreases by 3. Thus, the gradient tells us how fast $y$ changes relative to $x$.

Trigonometric Interpretation of Gradient

A more sophisticated way to view the gradient is as the tangent of the angle $\theta$ that the line makes with the positive $x$-axis (assuming equal scales on both axes). Therefore, $m = \tan \theta$. For example, a line at 45 degrees has $m = \tan 45^{\circ} = 1$, and a line at 135 degrees has $m = \tan 135^{\circ} = -1$.

Special Cases: Vertical and Horizontal Lines

Horizontal lines have a gradient of zero and take the form $y = k$, where $k$ is a constant. The $x$-axis itself is the line $y = 0$. Vertical lines do not have a strictly defined finite gradient (often called infinite) and take the form $x = k$. The $y$-axis is the line $x = 0$.

Parallel and Perpendicular Lines

Two lines $y = m_1x + c_1$ and $y = m_2x + c_2$ are parallel if and only if they have the same steepness, meaning $m_1 = m_2$.

Two lines are perpendicular if and only if the product of their gradients is -1, expressed as $m_1m_2 = -1$. This condition excludes horizontal and vertical lines, which are perpendicular but have gradients of 0 and undefined respectively. You can visualize this using similar triangles:

Forms of the Equation

In the TMUA, you must be comfortable with various algebraic forms of a line.

1. Point-Gradient Form: $y - y_1 = m(x - x_1)$. This is particularly useful when you know the gradient $m$ and a specific point $(x_1, y_1)$ on the line.
2. General Form: $ax + by + c = 0$. In this form, $a$, $b$, and $c$ are constants. Note that $c$ here is NOT the $y$-intercept. You should be able to rearrange this into $y = mx + c$ to identify the gradient.

Finding Equations of Straight Lines

To uniquely specify a line, you need two pieces of information. There are two primary cases:

Case 1: One point and the gradient

Given a point $(x_1, y_1)$ and a gradient $m$, the equation is derived from the fact that the gradient between any general point $(x, y)$ and $(x_1, y_1)$ must be $m$:

$\frac{y - y_1}{x - x_1} = m$

Worked Example: Find the line with gradient $m = 4$ through $(3, 2)$. Using $y - y_1 = m(x - x_1)$: $y - 2 = 4(x - 3)$ $y - 2 = 4x - 12$ $y = 4x - 10$

Worked Example: Find the line with gradient $m = -2$ through $(-2, -4)$. $y - (-4) = -2(x - (-2))$ $y + 4 = -2(x + 2)$ $y + 4 = -2x - 4$ $y = -2x - 8$

Case 2: Two distinct points

Given $(x_1, y_1)$ and $(x_2, y_2)$, first calculate the gradient $m = \frac{y_1 - y_2}{x_1 - x_2}$. Ensure the order of subtraction is consistent for both $x$ and $y$ to avoid sign errors.

Worked Example: Find the line through $(-2, 5)$ and $(-4, -7)$. First, find $m$: $m = \frac{5 - (-7)}{-2 - (-4)} = \frac{12}{2} = 6$. Now use $m = 6$ and the point $(-2, 5)$: $y - 5 = 6(x - (-2))$ $y - 5 = 6x + 12$ $y = 6x + 17$

Worked Example: Find the line through $(3, -7)$ and $(8, -7)$. Since the $y$-values are the same, the gradient is $m = \frac{-7 - (-7)}{8 - 3} = 0$. This is a horizontal line with the equation $y = -7$.