Coordinate Geometry of Straight Lines

The Gradient of a Straight Line

The standard form of a straight line equation is $y = mx + c$, where $m$ represents the gradient and $c$ represents the $y$-intercept. The gradient is a measure of the steepness of the line. If $m$ is positive, the line slopes upwards from bottom left to top right. If $m$ is negative, the line slopes downwards from top left to bottom right. The greater the absolute value of $m$, the steeper the line.

We can define steepness more precisely as the vertical distance moved for every 1 unit moved horizontally. For instance, if the gradient is 2, moving 1 unit to the right requires moving 2 units up to stay on the line. If the gradient is -3, moving 1 unit to the right requires moving 3 units down.

Another interpretation is that the gradient represents a rate of change. A gradient of 2 means $y$ changes twice as fast as $x$. If $x$ increases by 5, $y$ must increase by 10. A negative gradient indicates that as $x$ increases, $y$ decreases.

Gradient as a Trigonometric Tangent

If the scales on the $x$ and $y$ axes are identical, the gradient $m$ is equal to the tangent of the angle $\theta$ that the line makes with the positive $x$-axis, such that $m = \tan \theta$.

For a line at 45 degrees, the gradient is $\tan 45^\circ = 1$, as seen in the line $y = x$. For an angle of 135 degrees, the gradient is $\tan 135^\circ = -1$, as seen in $y = -x$.

Special Cases: Horizontal and Vertical Lines

Horizontal lines have a gradient of zero and take the form $y = k$, where $k$ is a constant. Vertical lines do not have a defined gradient, though they are sometimes described as having an infinite gradient. They take the form $x = k$. Note specifically that the $x$-axis has the equation $y = 0$, and the $y$-axis has the equation $x = 0$.

Parallel and Perpendicular Lines

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

Two lines are perpendicular if and only if the product of their gradients is $-1$, expressed as $m_1m_2 = -1$. This relationship can be understood by looking at similar triangles formed by the lines.

You can also justify this using trigonometry. Since the angles made with the $x$-axis differ by 90 degrees for perpendicular lines, the gradients $m_1 = \tan \theta$ and $m_2 = \tan (\theta + 90^\circ) = -\cot \theta$ satisfy $m_1m_2 = \tan \theta \times (-\frac{1}{\tan \theta}) = -1$.

Graph Transformations of y = x

The equations of straight lines can be viewed as transformations of the parent function $y = x$. Consider how the following changes affect the graph:

1. $y = mx$: A vertical stretch or reflection depending on the value of $m$.
2. $y - y_1 = x$: A vertical translation by $y_1$.
3. $y = x - x_1$: A horizontal translation by $x_1$.
4. $y - y_1 = m(x - x_1)$: A combination of stretching and both horizontal and vertical translations.

Finding the Equation of a Line

You can uniquely determine a line's equation given two pieces of information.

Case 1: A point $(x_1, y_1)$ and the gradient $m$

Since the gradient between any point $(x, y)$ on the line and the given point $(x_1, y_1)$ must be $m$, we have: $\frac{y - y_1}{x - x_1} = m$ Rearranging gives the point-slope form: $y - y_1 = m(x - x_1)$. Alternatively, substitute the point into $y = mx + c$ to solve for $c$.

Examples:

1. For $m = 4$ and point $(3, 2)$: $y - 2 = 4(x - 3) \rightarrow y = 4x - 10$.
2. For $m = -5$ and point $(5, 3)$: $y - 3 = -5(x - 5) \rightarrow y = -5x + 28$.
3. For $m = -2$ and point $(-2, -4)$: $y - (-4) = -2(x - (-2)) \rightarrow y + 4 = -2x - 4 \rightarrow y = -2x - 8$.

Case 2: Two distinct points $(x_1, y_1)$ and $(x_2, y_2)$

You can set the gradient between a general point $(x, y)$ and $(x_1, y_1)$ equal to the gradient between the two known points: $\frac{y - y_1}{x - x_1} = \frac{y_1 - y_2}{x_1 - x_2}$

Alternatively, calculate $m = \frac{y_1 - y_2}{x_1 - x_2}$ first, then use the method from Case 1. Ensure the order of subtraction is consistent for $x$ and $y$ to avoid sign errors.

Examples:

1. Points $(0, 0)$ and $(2, 3)$: $m = \frac{3-0}{2-0} = 1.5$. Since it passes through the origin, $y = 1.5x$.
2. Points $(-2, 5)$ and $(-4, -7)$: $m = \frac{5 - (-7)}{-2 - (-4)} = \frac{12}{2} = 6$. Using $(-2, 5)$, $y - 5 = 6(x + 2) \rightarrow y = 6x + 17$.
3. Points $(3, -7)$ and $(8, -7)$: The $y$-values are identical, so it is a horizontal line $y = -7$ with gradient 0.

The General Form ax + by + c = 0

Lines are frequently written as $ax + by + c = 0$. It is vital to note that the $c$ in this form is not the $y$-intercept. To find the gradient and intercept, rearrange the equation into the form $y = -\frac{a}{b}x - \frac{c}{b}$.