1974 AHSME Problems/Problem 11

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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