Difference between revisions of "1968 AHSME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | The original line can be adjusted to [[standard form]], in which it is <math>x-3y=7</math>. Because this line is in standard form, its slope is <math>\frac{-1}{-3}=\frac{1}{3}</math>. Thus, the slope of the perpendicular line is the negative reciprocal of this number, <math>-3</math>. Because we know that this new line passes through the point <math>(0,4)</math>, we can describe this line using [[point-slope form]], in which it is <math>y-4=-3(x-0)</math>, or <math>y+3x-4</math>, which is answer choice <math>\fbox{A}</math>. |
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=2|num-a=4}} | + | {{AHSME 35p box|year=1968|num-b=2|num-a=4}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:51, 17 July 2024
Problem
A straight line passing through the point 
is perpendicular to the line 
. Its equation is:
Solution
The original line can be adjusted to standard form, in which it is 
. Because this line is in standard form, its slope is 
. Thus, the slope of the perpendicular line is the negative reciprocal of this number, 
. Because we know that this new line passes through the point , we can describe this line using point-slope form, in which it is 
, or 
, which is answer choice 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
