Difference between revisions of "1974 AHSME Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | Notice that | + | Notice that since <math> (a, b) </math> is on <math> y=mx+k </math>, we have <math> b=am+k </math>. Similarly, <math> d=cm+k </math>. Using the distance formula, the distance between the points <math> (a, b) </math> and <math> (c, d) </math> is |
<math> \sqrt{(a-c)^2+(b-d)^2}=\sqrt{(a-c)^2+(am+k-cm-k)^2} </math> | <math> \sqrt{(a-c)^2+(b-d)^2}=\sqrt{(a-c)^2+(am+k-cm-k)^2} </math> | ||
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==See Also== | ==See Also== | ||
{{AHSME box|year=1974|num-b=10|num-a=12}} | {{AHSME box|year=1974|num-b=10|num-a=12}} | ||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 05:04, 21 August 2023
Problem
If and
are two points on the line whose equation is
, then the distance between
and
, in terms of
and
is
Solution
Notice that since is on
, we have
. Similarly,
. Using the distance formula, the distance between the points
and
is
And so the answer is .
See Also
1974 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.