Difference between revisions of "1992 AHSME Problems/Problem 12"
(→Solution) |
Math Kirby (talk | contribs) m |
||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | + | ||
First we want to put this is slope-intercept form, so we get <math>y=\dfrac{1}{3}x+\dfrac{11}{3}</math>. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since <math>m+b</math> is the sum of the slope and the y-intercept, we get <math>-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{-4}</math>. | First we want to put this is slope-intercept form, so we get <math>y=\dfrac{1}{3}x+\dfrac{11}{3}</math>. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since <math>m+b</math> is the sum of the slope and the y-intercept, we get <math>-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{-4}</math>. | ||
Latest revision as of 10:36, 30 October 2024
Problem
Let 
be the image when the line 
is reflected across the 
-axis. The value of 
is
Solution
First we want to put this is slope-intercept form, so we get 
. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since is the sum of the slope and the y-intercept, we get 
.
See also
| 1992 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
