1997 AHSME Problems/Problem 12
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[hide]Problem
If and are real numbers and , then the line whose equation is cannot contain the point
Solution 1
Geometrically, is the y-intercept, and is the slope.
Since , 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 for a positve , you can't go down into the fourth quadrant with a positive slope. Thus, point is a possible point.
Lines with a negative y-intercept and negative slope can go through quadrants II, III, and IV. Thus, point is a possible point.
Looking at the axes, any point on the y-axis is possible. Thus, and 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, cannot be reached.
Solution 2
Algebraically, if , then either and , or and .
Constructing lines that hit some of the points:
will hit , so is possible.
will hit , so is possible.
will hit , so is possible.
will hit , so is possible.
By process of elimination, must be impossible. Plugging in the point into will give , or . Thus, and must be of opposite signs.
Solution 3
Plugging in into gives . Thus, is positive, and can be anything, so can be positive too.
Plugging in into gives . Thus, is negative, and can be anything, so can be negative too.
Plugging in into gives . Examining this shows and can both be positive.
Plugging in into gives . Examining this shows and can both be negative.
Plugging in into gives . Examining this shows exactly one of or can be positive - the other must be negative. Thus, the answer is .
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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