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1997 AHSME Problems/Problem 12

Revision as of 08:32, 9 August 2011 by Talkinaway (talk | contribs)

Problem

If $m$ and $b$ are real numbers and $mb>0$, then the line whose equation is $y=mx+b$ cannot contain the point

$\textbf{(A)}\ (0,1997)\qquad\textbf{(B)}\ (0,-1997)\qquad\textbf{(C)}\ (19,97)\qquad\textbf{(D)}\ (19,-97)\qquad\textbf{(E)}\ (1997,0)$

Solution 1

Geometrically, $b$ is the y-intercept, and $m$ is the slope.

Since $mb>0$, then either we have a positive y-intercept and a positive slope, or a negative y-intercept and a negative slope.

Lines with a positive y-intercept and positive slope can go through quadrants I, II, and III. They cannot go through quadrant IV, because if you start at $(0,b)$ for a positve $b$, you can't go down into the fourth quadrant with a positive slope. Thus, point $C$ is a possible point.

Lines with a negative y-intercept and negative slope can go through quadrants II, III, and IV. Thus, point $D$ is a possible point.

Looking at the axes, any point on the y-axis is possible. Thus, $A$ and $B$ are both possible.

However, points on the positive x-axis are inpossible to reach. If you start with a positive y-intercept, you must go up and to the right. If you start with a negative y-intercept, you must go down and to the right. Thus, $\boxed{E}$ cannot be reached.

Solution 2

Algebraically, if $mb > 0$, then either $m>0$ and $b>0$, or $m<0$ and $b<0$.

Constructing lines that hit some of the points:

$y = x + 1997$ will hit $(0,1997)$, so $A$ is possible.

$y = -x - 1997$ will hit $(0,-1997)$, so $B$ is possible.

$y = x + 78$ will hit $(19,97)$, so $C$ is possible.

$y = -x - 78$ will hit $(19, -97)$, so $D$ is possible.

By process of elimination, $\boxed{E}$ must be impossible. Plugging in the point $(1997,0)$ into $y = mx + b$ will give $0 = 1997m + b$, or $b = -1997m$. Thus, $m$ and $b$ must be of opposite signs.

Solution 3

Plugging in $(0,1997)$ into $y = mx + b$ gives $1997 = 0m + b$. Thus, $b$ is positive, and $m$ can be anything, so $m$ can be positive too.

Plugging in $(0, -1997)$ into $y = mx + b$ gives $-1997 = 0m + b$. Thus, $b$ is negative, and $m$ can be anything, so $m$ can be negative too.

Plugging in $(19,97)$ into $y = mx + b$ gives $97 = 19m + b$. Examining this shows $m$ and $b$ can both be positive.

Plugging in $(19,-97)$ into $y = mx + b$ gives $-97 = 19m + b$. Examining this shows $m$ and $b$ can both be negative.

Plugging in $(1997,0)$ into $y = mx + b$ gives $0 = 1997m + b$. Examining this shows exactly one of $m$ or $b$ can be positive - the other must be negative. Thus, the answer is $\boxed{E}$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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