Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1997 AHSME Problems/Problem 12

Revision as of 21:46, 8 August 2011 by Talkinaway (talk | contribs) (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...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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, $E$ 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 $\boxed{E}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_12&oldid=41299"