Deriving the Equation of a Straight Line with a Given Gradient and Point

In this article, we will explore the process of finding the equation of a straight line given its gradient and a point through which it passes. We'll also discuss the different methods and their underlying principles, ensuring that the derived equation accurately represents the line.

The Point-Slope Form of a Line

The point-slope form of the equation of a line is:

y - y_1 m(x - x_1)

In this equation, (m) represents the gradient (or slope) of the line, and ((x_1, y_1)) is a point on the line. Given that the gradient (m -3) and the point is ((-3, 5)), we can substitute these values into the equation to find the line's equation.

Step-by-Step Derivation

Method 1: Direct Substitution

Substitute (m -3), (x_1 -3), and (y_1 5) into the point-slope form:

y - 5 -3(x - (-3))

Simplify the equation by distributing the (-3):

y - 5 -3x - 9

Add 5 to both sides to isolate (y):

y -3x - 4

Thus, the equation of the straight line is:

y -3x - 4

Method 2: Simplification and Interpretation

Given (m -3), we can express it as (-frac{9}{3}). Using this, we can rewrite and simplify the equation:

y - 5 -3(x 3)

Distribute (-3) to get:

y - 5 -3x - 9

Add 5 to both sides:

y -3x - 4

Thus, the equation of the straight line is:

y -3x - 4

Method 3: Using the Slope-Intercept Form

To derive the equation in the slope-intercept form (y mx c), we need to find the y-intercept (c). Starting from the point ((-3, 5)) and using the gradient, we can calculate:

y - 5 -3(x 3)

Expanding and simplifying:

y - 5 -3x - 9

Adding 5 to both sides:

y -3x - 4

Thus, the equation of the straight line is:

y -3x - 4

Understanding the Derived Equation

On a line with a slope of (-3), for every 1 unit increase in the (x) coordinate, there is a 3-unit decrease in the (y) coordinate. This is because the slope (-3) can be interpreted as a change in (y) over a change in (x), or (frac{Delta y}{Delta x} -3).

Using the point ((-3, 5)) and the slope (-3), we can verify the equation:

Step 1: Find the change in (x) to reach the y-axis (i.e., (x 0)). Since (x_1 -3) and the slope is (-3), the change in (x) is 3 (run).

Step 2: Calculate the change in (y) using the slope. The change in (y) is (5 - 3 -4) (rise).

Thus, the y-intercept is (-4), and the equation of the line is:

y -3x - 4

Conclusion

The equation of the straight line that has a gradient of (-3) and passes through the point ((-3, 5)) is:

y -3x - 4

This method can be applied to derive similar equations for other lines with given gradients and points. Understanding the principles behind the point-slope and slope-intercept forms of linear equations is essential for effectively solving such problems.