Introduction

In mathematics, understanding the concept of gradient and equations of straight lines is fundamental to various applications in algebra, geometry, and calculus. The gradient of a straight line represents its steepness or slope, while the equation of a straight line helps us to define and identify the relationship between variables. In this topic, we will explore how to calculate the gradient of a line, determine equations of straight lines in different forms, and solve problems related to them.

Gradient

The gradient of a straight line is defined as the ratio of the vertical change to the horizontal change between any two points on the line. It is denoted by the symbol $m$.

Key Terms:

• Gradient ($m$): The slope of a line, given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ where $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line.

Example:

Given two points $A(3, 5)$ and $B(7, 11)$, find the gradient of the line passing through these points.

Solution: Using the formula for gradient, we have: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 5}{7 - 3} = \frac{6}{4} = 1.5$ Therefore, the gradient of the line passing through points $A$ and $B$ is $1.5$.

Equation of a Straight Line

The equation of a straight line can be represented in different forms, such as slope-intercept form, point-slope form, and general form. These different forms provide valuable information about the line's characteristics.

Key Terms:

• Slope-Intercept Form: $y = mx + c$, where $m$ is the gradient and $c$ is the y-intercept.
• Point-Slope Form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line.
• General Form: $Ax + By + C = 0$, where $A$, $B$, and $C$ are constants.

Example:

Find the equation of a line with gradient $2$ passing through the point $(-3, 4)$ in slope-intercept form.

Solution: Using the point-slope form with $m = 2$ and $(-3, 4)$, we get: $y - 4 = 2(x + 3)$ $y - 4 = 2x + 6$ $y = 2x + 10$ Therefore, the equation of the line in slope-intercept form is $y = 2x + 10$.

Common Mistakes

• Misinterpreting the negative sign in front of the gradient as a subtraction operation.
• Forgetting to simplify fractions when calculating the gradient.
• Confusing the different forms of the equation of a straight line.

Key Points

• The gradient of a line indicates how steep the line is.
• The equation of a straight line can be represented in various forms like slope-intercept, point-slope, and general form.
• Understanding the relationship between gradients and equations of straight lines is crucial for solving problems in coordinate geometry.

Practice Questions

1. Find the gradient of a line passing through points $P(2, 3)$ and $Q(-1, 7)$.

   Solution: $m = \frac{7 - 3}{-1 - 2} = \frac{4}{-3} = -\frac{4}{3}$

2. Determine the equation of a line with gradient $-\frac{1}{2}$ passing through the point $(4, -3)$ in point-slope form.

   Solution: $y - (-3) = -\frac{1}{2}(x - 4)$ $y + 3 = -\frac{1}{2}x + 2$ $y = -\frac{1}{2}x - 1$

3. Write the equation of a line with gradient $3$ and y-intercept $-2$ in slope-intercept form.

   Solution: $y = 3x - 2$

4. Given the line $2x + 3y = 6$, rewrite the equation in slope-intercept form.

   Solution: $3y = -2x + 6$ $y = -\frac{2}{3}x + 2$

5. Calculate the gradient of a line parallel to the line $y = 4x + 1$.

   Solution: The gradient of a line parallel to $y = 4x + 1$ is $4$.

6. Find the equation of a line perpendicular to $2x + 3y = 5$ passing through the point $(-1, 4)$.

   Solution: The gradient of the perpendicular line is $\frac{-2}{3}$. Using point-slope form, $y - 4 = -\frac{2}{3}(x + 1)$ $3y - 12 = -2x - 2$ $2x + 3y = 10$