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16 Sep 2014, 00:44
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If the line \(y = kx + b\) intersects the x-axis at \((2, 0)\) and the y-axis at \((0, 5)\), what is the value of \(\frac{k}{b}\)?
A. \(-\frac{5}{2}\)
B. \(-\frac{5}{4}\)
C. \(-\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{4}{5}\)
A. \(-\frac{5}{2}\)
B. \(-\frac{5}{4}\)
C. \(-\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{4}{5}\)
16 Sep 2014, 00:44
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Official Solution:
If the line \(y = kx + b\) intersects the x-axis at \((2, 0)\) and the y-axis at \((0, 5)\), what is the value of \(\frac{k}{b}\)?
A. \(-\frac{5}{2}\)
B. \(-\frac{5}{4}\)
C. \(-\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{4}{5}\)
Substituting \((0, 5)\) for \((x, y)\) into \(y = kx + b\) gives us \(5 = 0 + b\), resulting in \(b = 5\).
Substituting \((2, 0)\) for \((x, y)\) into \(y = kx + b\) gives us \(0 = 2k + b\). Since we already know that \(b = 5\), then \(k = -\frac{5}{2}\).
Therefore, \(\frac{k}{b} = \frac{(-\frac{5}{2})}{5} = -\frac{1}{2}\).
Note that while Geometry is not tested on GMAT Focus, coordinate geometry is tested under the Functions and Graphing sections found in the Official Guide for GMAT Focus Edition.
Answer: C
If the line \(y = kx + b\) intersects the x-axis at \((2, 0)\) and the y-axis at \((0, 5)\), what is the value of \(\frac{k}{b}\)?
A. \(-\frac{5}{2}\)
B. \(-\frac{5}{4}\)
C. \(-\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{4}{5}\)
Substituting \((0, 5)\) for \((x, y)\) into \(y = kx + b\) gives us \(5 = 0 + b\), resulting in \(b = 5\).
Substituting \((2, 0)\) for \((x, y)\) into \(y = kx + b\) gives us \(0 = 2k + b\). Since we already know that \(b = 5\), then \(k = -\frac{5}{2}\).
Therefore, \(\frac{k}{b} = \frac{(-\frac{5}{2})}{5} = -\frac{1}{2}\).
Note that while Geometry is not tested on GMAT Focus, coordinate geometry is tested under the Functions and Graphing sections found in the Official Guide for GMAT Focus Edition.
Answer: C
29 Jan 2016, 14:35
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Alternate Solution :
Point 1 = (2,0)
Substituting for x and y in the first equation gives 0 = k(2) + b => 2k = -b => (k/b) = (-1/2)
IMO, we don't need the 2nd equation or the second point to solve this question.
Point 1 = (2,0)
Substituting for x and y in the first equation gives 0 = k(2) + b => 2k = -b => (k/b) = (-1/2)
IMO, we don't need the 2nd equation or the second point to solve this question.
