Point-gradient equation

Topic summary

The point-gradient equation of a straight line is a formula used to write the equation of a line when you know the gradient \( m \) and a point \( (x_1, y_1) \) on the line. The general form of the point-gradient equation is:

\[ y – y_1 = m(x – x_1) \]

Where:

• \( m \) is the gradient (or slope) of the line.
• \( (x_1, y_1) \) is a known point on the line.
• \( x \) and \( y \) are the variables representing any point on the line.

This formula is particularly useful when you know the gradient and a single point, as it allows you to quickly find the equation of the line.

1. Understanding the point-gradient equation:

The point-gradient equation is derived from the general equation \( y = mx + c \), but it is written in a form that is more convenient when a specific point on the line is known. It expresses the relationship between the gradient and a point on the line. By using the known point, we can easily substitute its values to find the equation of the line.

2. Steps to use the point-gradient equation:

To find the equation of the line using the point-gradient equation, follow these steps:

• Write down the point \( (x_1, y_1) \) and the gradient \( m \) of the line.
• Substitute the values of \( m \), \( x_1 \), and \( y_1 \) into the point-gradient equation: \[ y – y_1 = m(x – x_1) \]
• Simplify the equation as needed to get it into the desired form (e.g., slope-intercept form or general form).

3. Example: Using the point-gradient equation:

Let’s find the equation of a line with gradient \( m = 3 \) that passes through the point \( (2, 5) \).

Step 1: Write the point-gradient equation:

Substitute \( m = 3 \), \( x_1 = 2 \), and \( y_1 = 5 \) into the point-gradient equation: \[ y – 5 = 3(x – 2) \]

Step 2: Simplify the equation:

Distribute the \( 3 \) on the right-hand side: \[ y – 5 = 3x – 6 \]

Step 3: Isolate \( y \):

Add \( 5 \) to both sides to solve for \( y \): \[ y = 3x – 6 + 5 \] \[ y = 3x – 1 \]

4. The equation of the line:

The equation of the line is: \[ y = 3x – 1 \]

5. Check the solution:

To verify, substitute \( x = 2 \) into the equation to ensure the point \( (2, 5) \) satisfies the equation: \[ y = 3(2) – 1 = 6 – 1 = 5 \] This is correct, so the equation of the line is verified as \( y = 3x – 1 \).

6. Special cases to note:

• If the gradient \( m = 0 \), the line is horizontal, and the equation becomes: \[ y – y_1 = 0(x – x_1) \quad \text{or simply} \quad y = y_1 \]
• If the gradient \( m \) is undefined (i.e., the line is vertical), the equation becomes: \[ x = x_1 \]