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In the \(\mathrm{xy}\text{-plane}\), line k has positive slope and \(\mathrm{x}\text{-intercept } 4\). If the area of the triangle formed by line k and the two axes is 12, what is the \(\mathrm{y}\text{-intercept}\) of line k ?
Let's break down what we know in simple terms. We have a line called 'k' that crosses the x-axis at the point \((4, 0)\) - this means when y equals zero, x equals 4. We also know this line has a positive slope, which means it goes upward as we move from left to right.
The line also crosses the y-axis somewhere - that's what we're trying to find. When a line crosses the y-axis, x equals zero at that point.
Finally, we know that this line, together with the x-axis and y-axis, forms a triangle with an area of 12 square units.
Process Skill: TRANSLATE - Converting the problem's geometric language into clear mathematical understanding
Let's picture what's happening here. Imagine the x-axis and y-axis forming a big plus sign. Our line k crosses the x-axis at point \((4, 0)\) and crosses the y-axis at some unknown point \((0, b)\) where b is our y-intercept.
Since the line has positive slope and crosses the x-axis at \((4, 0)\), think about where it must cross the y-axis. A line with positive slope going through \((4, 0)\) would have to cross the y-axis below the origin (at a negative y-value) to maintain that positive slope.
The triangle is formed by:
- The segment of the x-axis from the origin \((0, 0)\) to the x-intercept \((4, 0)\)
- The segment of the y-axis from the origin \((0, 0)\) to the y-intercept \((0, b)\)
- The line k connecting these two intercept points
Process Skill: VISUALIZE - Understanding the geometric setup is crucial for applying the area formula correctly
Now here's a key insight about triangles formed by lines and the coordinate axes: the area of such a triangle equals half the product of the absolute values of the intercepts.
Think of it this way: our triangle has a base along the x-axis with length 4 (from 0 to 4), and a height along the y-axis with length \(|b|\) (the absolute value of our y-intercept).
Using the basic triangle area formula: \(\mathrm{Area} = \frac{1}{2} \times \mathrm{base} \times \mathrm{height}\)
So: \(\mathrm{Area} = \frac{1}{2} \times 4 \times |b| = 2|b|\)
We know the area is 12, so:
\(12 = 2|b|\)
\(|b| = 6\)
This means \(b = 6\) or \(b = -6\).
Now we need to determine which value is correct: \(b = 6\) or \(b = -6\).
Remember our constraint: the line has positive slope. Let's think about what this means:
- If the y-intercept were +6, the line would go from \((0, 6)\) to \((4, 0)\)
- The slope would be \(\frac{0 - 6}{4 - 0} = \frac{-6}{4} = -\frac{3}{2}\), which is negative
- If the y-intercept is -6, the line would go from \((0, -6)\) to \((4, 0)\)
- The slope would be \(\frac{0 - (-6)}{4 - 0} = \frac{6}{4} = \frac{3}{2}\), which is positive
Since we need positive slope, the y-intercept must be -6.
Process Skill: APPLY CONSTRAINTS - Using the positive slope requirement to eliminate the wrong solution
The y-intercept of line k is -6.
Verification: A line from \((0, -6)\) to \((4, 0)\) has slope \(\frac{3}{2}\) (positive ✓), forms a triangle with area \(\frac{1}{2} \times 4 \times 6 = 12\) (correct ✓), and has x-intercept 4 (given ✓).
The answer is A. -6
1. Misunderstanding the triangle formation: Students often struggle to visualize which triangle is being formed. They might think the triangle is formed by the line and two coordinate axes in a different configuration, or fail to recognize that the vertices are at the origin \((0,0)\), the x-intercept \((4,0)\), and the y-intercept \((0,b)\).
2. Ignoring the positive slope constraint: Students may focus solely on the area calculation and forget that the line must have a positive slope. This constraint is crucial for determining which of the two possible y-intercept values is correct.
3. Confusion about triangle area formula application: Students might not realize that for a triangle formed by a line and the coordinate axes, the area formula simplifies to \(\frac{1}{2} \times |\mathrm{x-intercept}| \times |\mathrm{y-intercept}|\). They may try to use more complex geometric approaches or struggle with setting up the area calculation correctly.
1. Arithmetic errors in slope calculation: When checking which y-intercept value gives a positive slope, students often make sign errors. For example, when calculating the slope from \((0,-6)\) to \((4,0)\), they might compute \(\frac{0-(-6)}{4-0}\) incorrectly as \(-\frac{6}{4}\) instead of \(+\frac{6}{4}\).
2. Forgetting absolute value in area calculation: Students may write the area equation as \(12 = 2b\) instead of \(12 = 2|b|\), leading them to conclude that \(b = 6\) without considering the negative possibility. This prevents them from properly evaluating both potential solutions.
3. Sign confusion with intercepts: When working with negative y-intercepts, students often get confused about signs, especially when substituting into formulas or when moving between the concepts of 'y-intercept value' and 'distance from origin'.
1. Selecting the positive y-intercept value: After finding that \(|b| = 6\), students might automatically choose \(b = 6\) (answer choice E) without checking the slope constraint. This is a common error because positive values often 'feel' more natural as answers.
2. Failing to verify the final answer: Students may not double-check that their chosen y-intercept actually satisfies all the given conditions (positive slope, correct area, and given x-intercept). This verification step would catch errors made in earlier phases.