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Difference between revisions of "1974 AHSME Problems/Problem 11"
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==Solution==
 
==Solution==
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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
+
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>  

Latest revision as of 04:04, 21 August 2023

Problem

If $(a, b)$ and $(c, d)$ are two points on the line whose equation is $y=mx+k$, then the distance between $(a, b)$ and $(c, d)$, in terms of $a, c,$ and $m$ is

$\mathrm{(A)\ } |a-c|\sqrt{1+m^2} \qquad \mathrm{(B) \ }|a+c|\sqrt{1+m^2} \qquad \mathrm{(C) \ } \frac{|a-c|}{\sqrt{1+m^2}} \qquad$

$\mathrm{(D) \ } |a-c|(1+m^2) \qquad \mathrm{(E) \ }|a-c|\,|m|$

Solution

Notice that since $(a, b)$ is on $y=mx+k$, we have $b=am+k$. Similarly, $d=cm+k$. Using the distance formula, the distance between the points $(a, b)$ and $(c, d)$ is

$\sqrt{(a-c)^2+(b-d)^2}=\sqrt{(a-c)^2+(am+k-cm-k)^2}$

$=\sqrt{(a-c)^2+m^2(a-c)^2}$

$=|a-c|\sqrt{1+m^2}.$

And so the answer is $\boxed{\text{A}}$.

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

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