Introduction

In the realm of mathematics, understanding linear equations is fundamental. This article delves into a specific problem involving a straight line, where the gradient is given. By exploring the step-by-step solution, you will gain a deeper understanding of how to work with linear equations.

Solving for k in a Straight Line Problem

Given two points on a straight line, (3k, 6) and (k, -5), with a gradient of 2, the goal is to find the value of k. We can use the formula for the gradient (slope) of a line that passes through two points (x1, y1) and (x2, y2): m (y2 - y1)/(x2 - x1).

Setting Up the Equation

Let's assign the points as follows:

Given that the gradient m is 2, we can set up the equation:

2 (y2 - y1)/(x2 - x1) (-5 - 6)/(k - 3k)

Simplifying the Equation

Substitute the given points into the equation:

2 (-5 - 6)/(k - 3k)

This simplifies to:

2 (-11)/(-2k)

We can then further simplify:

2 11/(2k)

Solving for k

Now, cross-multiply to solve for k:

2 * 2k 11

4k 11

k 11/4

Conclusion

Therefore, the value of k is boxed{11/4}, or 2.75 in decimal form.

Key Concepts

Understanding the gradient (or slope) of a line is crucial. The formula for the gradient of a line passing through two points is essential:

m (y2 - y1)/(x2 - x1)

By applying this formula, you can solve for any unknown variable in the context of linear equations.