A line in the xy-plane has a slope of -{1/4} and an x-intercept of 12. Which equation represents this line?
GMAT Algebra : (Alg) Questions
A line in the xy-plane has a slope of \(-\frac{1}{4}\) and an x-intercept of \(12\). Which equation represents this line?
\(\mathrm{y = \left(-\frac{1}{4}\right)x - 3}\)
\(\mathrm{y = \left(-\frac{1}{4}\right)x + 3}\)
\(\mathrm{y = \frac{1}{4}x - 3}\)
\(\mathrm{y = \frac{1}{4}x + 3}\)
\(\mathrm{y = \left(-\frac{1}{4}\right)x + 12}\)
1. TRANSLATE the problem information
- Given information:
- Slope = \(\mathrm{-\frac{1}{4}}\)
- x-intercept = \(\mathrm{12}\)
- What this tells us: The line passes through the point \(\mathrm{(12, 0)}\) because x-intercept means \(\mathrm{y = 0}\) when \(\mathrm{x = 12}\)
2. INFER the approach
- We need to find the equation of the line in slope-intercept form: \(\mathrm{y = mx + b}\)
- We know \(\mathrm{m = -\frac{1}{4}}\), but we need to find \(\mathrm{b}\) (the y-intercept)
- Since we have a point on the line \(\mathrm{(12, 0)}\) and the slope, we can substitute these values to solve for \(\mathrm{b}\)
3. SIMPLIFY to find the y-intercept
- Substitute the point \(\mathrm{(12, 0)}\) and slope into \(\mathrm{y = mx + b}\):
\(\mathrm{0 = (-\frac{1}{4})(12) + b}\) - Calculate:
\(\mathrm{0 = -3 + b}\) - Solve for \(\mathrm{b}\):
\(\mathrm{b = 3}\)
4. Write the final equation
- Now we have \(\mathrm{m = -\frac{1}{4}}\) and \(\mathrm{b = 3}\)
- The equation is \(\mathrm{y = (-\frac{1}{4})x + 3}\)
Answer: B
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students confuse x-intercept with y-intercept and think the line passes through \(\mathrm{(0, 12)}\) instead of \(\mathrm{(12, 0)}\).
When they substitute \(\mathrm{(0, 12)}\) into \(\mathrm{y = mx + b}\):
\(\mathrm{12 = (-\frac{1}{4})(0) + b}\)
\(\mathrm{12 = 0 + b}\)
\(\mathrm{b = 12}\)
This leads them to the equation \(\mathrm{y = (-\frac{1}{4})x + 12}\), causing them to select Choice E.
Second Most Common Error:
Poor SIMPLIFY execution: Students make sign errors when calculating \(\mathrm{(-\frac{1}{4})(12)}\).
They might calculate \(\mathrm{(-\frac{1}{4})(12) = +3}\) instead of \(\mathrm{-3}\), leading to:
\(\mathrm{0 = 3 + b}\)
\(\mathrm{b = -3}\)
This gives them \(\mathrm{y = (-\frac{1}{4})x - 3}\), causing them to select Choice A.
The Bottom Line:
The key challenge is correctly interpreting what "x-intercept of 12" means geometrically and then executing the algebra without sign errors. Students who master the TRANSLATE skill of converting intercepts to coordinate points will find this problem straightforward.
\(\mathrm{y = \left(-\frac{1}{4}\right)x - 3}\)
\(\mathrm{y = \left(-\frac{1}{4}\right)x + 3}\)
\(\mathrm{y = \frac{1}{4}x - 3}\)
\(\mathrm{y = \frac{1}{4}x + 3}\)
\(\mathrm{y = \left(-\frac{1}{4}\right)x + 12}\)