- Geometric analysis of the number line shows that if \(b\) is at the same distance from 0 as \(a\) but in the opposite direction, the relationship is \(b = -a\).
- The definition of direct variation, \(b = ka\), is satisfied by \(b = -a\) with \(k = -1\).
- Verification with the example \(b = 2 \frac{3}{4}\) and \(a = -2 \frac{3}{4}\) confirms that \(b = -a\) holds true.
- Final answer: \(\boxed{b = -a}\).
Explanation
- Analyzing the relationship between \(a\) and \(b\)
We are given that a number \(b\) is located at the same distance from 0 as a number \(a\), but in the opposite direction. On a number line, the distance of a point from zero is given by its absolute value. If \(b\) and \(a\) are the same distance from zero, then \(|b| = |a|\). Since they are in opposite directions, it must be that \(b = -a\). We are also told that \(b\) varies directly with \(a\). The definition of direct variation states that \(b = k \cdot a\) for some constant \(k\). Substituting \(b = -a\) into this form, we find that \(k = -1\), which satisfies the condition.
- Evaluating the given options and verifying the example
Let's verify the given example: \(b = 2 \frac{3}{4}\) when \(a = -2 \frac{3}{4}\). Converting these to improper fractions, we have \(b = \frac{11}{4}\) and \(a = -\frac{11}{4}\). Testing the relationship \(b = -a\), we get:
\[ \frac{11}{4} = -(-\frac{11}{4}) = \frac{11}{4}. \]
This confirms that the relationship \(b = -a\) is consistent with the provided example. Now, let's examine the other options:
- \(-b = -a\) simplifies to \(b = a\), which means they are in the same direction, contradicting the problem description.
- \(b - a = 0\) simplifies to \(b = a\), which similarly contradicts the description.
Thus, the only valid equation is \(b = -a\).
Examples
Direct variation is a fundamental concept used in various real-life scenarios, such as converting between different units or calculating costs. For example, if you are buying apples that cost \$2 per pound, the total cost \(C\) varies directly with the weight \(w\) in pounds, represented by the equation \(C = 2w\). In this problem, the relationship \(b = -a\) acts as a linear transformation where the constant of variation is \(k = -1\), showing how one value scales perfectly with another. This is similar to how a mirror image on a coordinate plane behaves, where every point \((x, y)\) corresponds to another point \((-x, -y)\) through the origin.