Equation of a Line Revision Notes for Grade 11 NSC Matric Mathematics

Equation of a Line

Understanding how to find the equation of a straight line is essential in analytical geometry. There are three main forms of line equations, each useful depending on the information you're given in a problem.

Different forms of line equations

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The form you choose depends on what information is provided in your problem. Each form has specific advantages that make calculations easier in different scenarios.

The form you choose depends on what information is provided:

Two-point form: Use when you know two points on the line
Gradient-point form: Use when you know the gradient and one point on the line
Gradient-intercept form: Use when you know the gradient and y-intercept, or as the final answer form

The two-point form of a line equation

When you have two points on a line, you can find the equation using the two-point form. This is particularly useful because any two distinct points determine exactly one straight line.

Formula: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

This formula works by setting up equal ratios between any point $(x, y)$ on the line and the two known points $(x_1, y_1)$ and $(x_2, y_2)$ .

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Worked Example: Finding the equation using two points

Question: Find the equation of the line passing through $P(-1; -5)$ and $Q(5; 4)$ .

Solution:

Step 1: Draw a sketch to visualise the problem

Step 2: Assign variables to the coordinates

Let $P(x_1; y_1) = (-1; -5)$ and $Q(x_2; y_2) = (5; 4)$

Therefore: $x_1 = -1$ , $y_1 = -5$ , $x_2 = 5$ , $y_2 = 4$

Step 3: Write the two-point form formula

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

Step 4: Substitute the values and solve for $y$

$\frac{y - (-5)}{x - (-1)} = \frac{4 - (-5)}{5 - (-1)}$

$\frac{y + 5}{x + 1} = \frac{9}{6} = \frac{3}{2}$

Cross multiply: $y + 5 = \frac{3}{2}(x + 1)$

$y + 5 = \frac{3}{2}x + \frac{3}{2}$

$y = \frac{3}{2}x + \frac{3}{2} - 5$

$y = \frac{3}{2}x - 3\frac{1}{2}$

Answer: $y = \frac{3}{2}x - 3\frac{1}{2}$

The gradient-point form of a line equation

The gradient-point form is derived from the two-point form by substituting the gradient formula. This form is useful when you know the gradient and coordinates of one point on the line.

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

where $m$ is the gradient and $(x_1, y_1)$ is the known point.

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Remember that the gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$ represents the "rise over run" - how much $y$ changes for each unit change in $x$ .

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Worked Example: Using gradient and one point

Question: Find the equation of the line with gradient $m = -\frac{1}{3}$ passing through the point $(-1; 1)$ .

Solution:

Step 1: Draw a sketch

Since $m < 0$ , the line has a negative slope (decreases as $x$ increases).

Step 2: Write the gradient-point form

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

Step 3: Substitute the known values

$y - 1 = -\frac{1}{3}(x - (-1))$

$y - 1 = -\frac{1}{3}(x + 1)$

$y - 1 = -\frac{1}{3}x - \frac{1}{3}$

$y = -\frac{1}{3}x - \frac{1}{3} + 1$

$y = -\frac{1}{3}x + \frac{2}{3}$

Answer: $y = -\frac{1}{3}x + \frac{2}{3}$

When you have two points, you can also use the gradient-point form by first calculating the gradient.

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Worked Example: Using gradient-point form with two points

Question: Find the equation of the line passing through $(-3; 2)$ and $(5; 8)$ .

Solution:

Step 1: Draw a sketch

Step 2: Assign variables to coordinates

$x_1 = -3$ , $y_1 = 2$ , $x_2 = 5$ , $y_2 = 8$

Step 3: Calculate the gradient

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{5 - (-3)} = \frac{6}{8} = \frac{3}{4}$

Step 4: Use gradient-point form with either point Using point $(-3; 2)$ : $y - y_1 = m(x - x_1)$

$y - 2 = \frac{3}{4}(x - (-3))$

$y - 2 = \frac{3}{4}(x + 3)$

$y - 2 = \frac{3}{4}x + \frac{9}{4}$

$y = \frac{3}{4}x + \frac{9}{4} + 2$

$y = \frac{3}{4}x + 4\frac{1}{4}$

Answer: $y = \frac{3}{4}x + 4\frac{1}{4}$

The gradient-intercept form of a line equation

The gradient-intercept form (also called the standard form) is the most commonly used form for expressing line equations.

Formula: $y = mx + c$

where:

$m$ is the gradient of the line
$c$ is the y-intercept (where the line crosses the y-axis)

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This form is derived from the gradient-point form by expanding and rearranging. When $x = 0$ , we get $y = c$ , confirming that $c$ is the y-intercept.

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Worked Example: Using gradient and y-intercept

Question: Find the equation of the line with gradient $m = -2$ passing through the point $(-1; 7)$ .

Solution:

Step 1: Understand the slope

Since $m < 0$ , the line decreases as x increases.

Step 2: Write the gradient-intercept form

$y = mx + c$

Step 3: Substitute the gradient

$y = -2x + c$

Step 4: Find $c$ using the given point

Substitute $(-1; 7)$ into the equation:

$7 = -2(-1) + c$

$7 = 2 + c$

$c = 5$

Step 5: Write the final equation

$y = -2x + 5$

The y-intercept is (0; 5).

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Worked Example: Finding equation from two points

Question: Find the equation of the line passing through $(-2; -7)$ and $(3; 8)$ .

Solution:

Step 1: Draw a sketch

Step 2: Set up the gradient-intercept form $y = mx + c$

Step 3: Use both points to create simultaneous equations

For point $(-2; -7)$ : $-7 = m(-2) + c$ → $-7 = -2m + c$ ... (1)

For point $(3; 8)$ : $8 = m(3) + c$ → $8 = 3m + c$ ... (2)

Step 4: Solve the simultaneous equations

Subtract equation (1) from equation (2):

$8 - (-7) = 3m + c - (-2m + c)$

$15 = 5m$

$m = 3$

Step 5: Find $c$ by substituting back

Using equation (2): $8 = 3(3) + c$

$8 = 9 + c$

$c = -1$

Answer: $y = 3x - 1$

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Exam Tips:

Always draw a sketch first to visualise the problem
Choose the most appropriate form based on the given information
When using two-point form, be careful with negative coordinates in calculations
Double-check your final answer by substituting the original points back into your equation
The gradient-intercept form $y = mx + c$ is often preferred as the final answer

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Key Points to Remember:

Two-point form $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$ is used when you have two points on the line
Gradient-point form $y - y_1 = m(x - x_1)$ is used when you know the gradient and one point
Gradient-intercept form $y = mx + c$ shows the gradient m and y-intercept c clearly
The gradient formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$ (rise over run)
All three forms represent the same line - choose the most convenient form for your given information

Equation of a Line (Grade 11 NSC Matric Mathematics): Revision Notes