https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_12&feed=atom&action=history1997 AHSME Problems/Problem 12 - Revision history2024-03-28T22:54:18ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_12&diff=56510&oldid=prevNathan wailes at 18:12, 5 July 20132013-07-05T18:12:50Z<p></p>
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</table>Nathan waileshttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_12&diff=41309&oldid=prevTalkinaway at 13:32, 9 August 20112011-08-09T13:32:11Z<p></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Plugging in <math>(1997,0)</math> into <math>y = mx + b</math> gives <math>0 = 1997m + b</math>.  Examining this shows exactly one of <math>m</math> or <math>b</math> can be positive - the other must be negative.  Thus, the answer is <math>\boxed{E}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Plugging in <math>(1997,0)</math> into <math>y = mx + b</math> gives <math>0 = 1997m + b</math>.  Examining this shows exactly one of <math>m</math> or <math>b</math> can be positive - the other must be negative.  Thus, the answer is <math>\boxed{E}</math>.</div></td></tr>
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</table>Talkinawayhttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_12&diff=41300&oldid=prevTalkinaway: /* Solution 2 */2011-08-09T03:00:29Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>y = -x - 78</math> will hit <math>(19, -97)</math>, so <math>D</math> is possible.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>y = -x - 78</math> will hit <math>(19, -97)</math>, so <math>D</math> is possible.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>By process of elimination, <math>E</math> must be impossible.  Plugging in the point <math>(1997,0)</math> into <math>y = mx + b</math> will give <math>0 = 1997m + b</math>, or <math>b = -1997m</math>.  Thus, <math>m</math> and <math>b</math> must be of opposite signs.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>By process of elimination, <math><ins class="diffchange diffchange-inline">\boxed{</ins>E<ins class="diffchange diffchange-inline">}</ins></math> must be impossible.  Plugging in the point <math>(1997,0)</math> into <math>y = mx + b</math> will give <math>0 = 1997m + b</math>, or <math>b = -1997m</math>.  Thus, <math>m</math> and <math>b</math> must be of opposite signs.</div></td></tr>
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</table>Talkinawayhttps://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_12&diff=41299&oldid=prevTalkinaway: Created page with "==Problem== If <math>m</math> and <math>b</math> are real numbers and <math>mb>0</math>, then the line whose equation is <math>y=mx+b</math> ''cannot'' contain the point <math..."2011-08-09T02:46:40Z<p>Created page with "==Problem== If <math>m</math> and <math>b</math> are real numbers and <math>mb>0</math>, then the line whose equation is <math>y=mx+b</math> ''cannot'' contain the point <math..."</p>
<p><b>New page</b></p><div>==Problem==<br />
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If <math>m</math> and <math>b</math> are real numbers and <math>mb>0</math>, then the line whose equation is <math>y=mx+b</math> ''cannot'' contain the point<br />
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<math> \textbf{(A)}\ (0,1997)\qquad\textbf{(B)}\ (0,-1997)\qquad\textbf{(C)}\ (19,97)\qquad\textbf{(D)}\ (19,-97)\qquad\textbf{(E)}\ (1997,0) </math><br />
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==Solution 1==<br />
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Geometrically, <math>b</math> is the y-intercept, and <math>m</math> is the slope.<br />
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Since <math>mb>0</math>, then either we have a positive y-intercept and a positive slope, or a negative y-intercept and a negative slope.<br />
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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 <math>(0,b)</math> for a positve <math>b</math>, you can't go down into the fourth quadrant with a positive slope. Thus, point <math>C</math> is a possible point.<br />
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Lines with a negative y-intercept and negative slope can go through quadrants II, III, and IV. Thus, point <math>D</math> is a possible point.<br />
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Looking at the axes, any point on the y-axis is possible. Thus, <math>A</math> and <math>B</math> are both possible.<br />
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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, <math>\boxed{E}</math> cannot be reached.<br />
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==Solution 2==<br />
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Algebraically, if <math>mb > 0</math>, then either <math>m>0</math> and <math>b>0</math>, or <math>m<0</math> and <math>b<0</math>.<br />
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Constructing lines that hit some of the points:<br />
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<math>y = x + 1997</math> will hit <math>(0,1997)</math>, so <math>A</math> is possible.<br />
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<math>y = -x - 1997</math> will hit <math>(0,-1997)</math>, so <math>B</math> is possible.<br />
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<math>y = x + 78</math> will hit <math>(19,97)</math>, so <math>C</math> is possible.<br />
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<math>y = -x - 78</math> will hit <math>(19, -97)</math>, so <math>D</math> is possible.<br />
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By process of elimination, <math>E</math> must be impossible. Plugging in the point <math>(1997,0)</math> into <math>y = mx + b</math> will give <math>0 = 1997m + b</math>, or <math>b = -1997m</math>. Thus, <math>m</math> and <math>b</math> must be of opposite signs.<br />
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==Solution 3==<br />
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Plugging in <math>(0,1997)</math> into <math>y = mx + b</math> gives <math>1997 = 0m + b</math>. Thus, <math>b</math> is positive, and <math>m</math> can be anything, so <math>m</math> can be positive too.<br />
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Plugging in <math>(0, -1997)</math> into <math>y = mx + b</math> gives <math>-1997 = 0m + b</math>. Thus, <math>b</math> is negative, and <math>m</math> can be anything, so <math>m</math> can be negative too.<br />
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Plugging in <math>(19,97)</math> into <math>y = mx + b</math> gives <math>97 = 19m + b</math>. Examining this shows <math>m</math> and <math>b</math> can both be positive.<br />
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Plugging in <math>(19,-97)</math> into <math>y = mx + b</math> gives <math>-97 = 19m + b</math>. Examining this shows <math>m</math> and <math>b</math> can both be negative.<br />
<br />
Plugging in <math>(1997,0)</math> into <math>y = mx + b</math> gives <math>0 = 1997m + b</math>. Examining this shows exactly one of <math>m</math> or <math>b</math> can be positive - the other must be negative. Thus, the answer is <math>\boxed{E}</math>.</div>Talkinaway