3.1.3 Equation of a Straight Line

1. Slope-Intercept Form:

The most common form of a straight line equation is the slope-intercept form:

$y = mx + c$

$m$ is the slope of the line, which measures its steepness. The slope $m$ is calculated as:

$m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$

where $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line.

$c$ is the $y$ -intercept, the point where the line crosses the $y$ -axis (when $x = 0$ ).

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Example:

Given the slope $m = 2$ and $y$ -intercept $c = -3$ , the equation of the line is:

$y = 2x - 3$

This line has a slope of $2$ and crosses the $y$ -axis at $(0, -3)$ .

2. Point-Slope Form:

If you know the slope $m$ of the line and one point $(x_1, y_1)$ on the line, you can use the point-slope form:

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

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Example:

Suppose you know a line passes through the point $(3, 4)$ and has a slope of $m = -2$ . The equation of the line is:

$y - 4 = -2(x - 3)$

Simplifying this:

$y = -2x + 6 + 4$

$y = -2x + 10$

3. General Form:

The equation of a line can also be written in the general form:

$Ax + By + C = 0$

$A$ , $B$ , and $C$ are constants.
This form is useful when dealing with lines in various algebraic contexts.

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Example:

The line $y = 2x - 3$ can be written in general form by rearranging the terms:

$2x - y - 3 = 0$

4. Two-Point Form:

If you know two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line, you can use the two-point form:

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

This is essentially the same as the point-slope form, but with the slope explicitly calculated using the two given points.

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Example:

Given points $(1, 2)$ and $(4, 8)$ , the equation of the line is:

$y - 2 = \frac{8 - 2}{4 - 1}(x - 1)$

$y - 2 = 2(x - 1)$

$y = 2x - 2 + 2$

$y = 2x$

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Example:

Deduce from rearrangement the gradient and y-intercept of $6x + 5y = 3$ : $5y = 3 - 6x$ $y = \frac{3}{5} - \frac{6}{5}x$
Gradient $m = -\frac{6}{5}$
$y$ -intercept $c = \frac{3}{5}$

Point-Slope Form:

The equation of a straight line can also be written as: $y - y_1 = m(x - x_1)$ where $(x_1, y_1)$ is any point on the line.

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Example:

Find the equation of the line that passes through $(2, 7)$ and has a gradient $m = 4$ : $y - 7 = 4(x - 2)$ $y - 7 = 4x - 8$

$y = 4x - 1$

The equation of the line in slope-intercept form is $\boxed {y = 4x - 1}$ .

Example: Finding the Equation of a Line

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Problem: Find the equation of the line that passes through $(4, 7)$ and $(9, -12)$ in the form $ax + by + c = 0$ , where $a, b, c \in \mathbb{Z}$ .

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Solution:

Calculate the Gradient: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-12 - 7}{9 - 14} = \frac{-19}{-5} = \frac{19}{5}$
Use the Point-Slope Form: $y - y_1 = m(x - x_1)$ Using the point $(4, 7)$ : $y - 7 = \frac{19}{5}(x - 14)$
Clear the Fraction: Multiply both sides by $5$ : $5(y - 7) = 19(x - 14)$ $5y - 35 = 19x - 266$
Rearrange to Standard Form: $5y - 35 = 19x - 266$ $5y - 19x + 231 = 0$
General Form: $-19x + 5y + 231 = 0$

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Notes:

We could use $(9, -12)$ instead of $(4, 7)$ , but using positive numbers simplifies calculations.
Multiplying both sides by 5 ensures all coefficients are integers.
Any integer multiple of the equation is also acceptable, e.g., $-1900x + 500y + 23100 = 0$ , but the simplest form is preferable.

Final Answer:

$\boxed {-19x + 5y + 231 = 0}$

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Summary:

Slope-Intercept Form: $y = mx + c$
Point-Slope Form: $y - y_1 = m(x - x_1)$
General Form: $Ax + By + C = 0$
Two-Point Form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of a Straight Line Simplified Revision Notes for A-Level AQA Maths Pure

3.1.3 Equation of a Straight Line

1. Slope-Intercept Form:

Example:

2. Point-Slope Form:

Example:

3. General Form:

Example:

4. Two-Point Form:

Example:

Example:

Point-Slope Form:

Example:

Example: Finding the Equation of a Line

Solution:

Notes:

Final Answer:

Summary:

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