1977 AHSME Problems/Problem 5

Problem 5

The set of all points $P$ such that the sum of the (undirected) distances from $P$ to two fixed points $A$ and $B$ equals the distance between $A$ and $B$ is

$\textbf{(A) }\text{the line segment from }A\text{ to }B\qquad \textbf{(B) }\text{the line passing through }A\text{ and }B\qquad\\ \textbf{(C) }\text{the perpendicular bisector of the line segment from }A\text{ to }B\qquad\\ \textbf{(D) }\text{an ellipse having positive area}\qquad \textbf{(E) }\text{a parabola}$


Solution by e_power_pi_times_i

The answer is $\textbf{(A)}$ because $P$ has to be on the line segment $AB$ in order to satisfy $PA+PB=AB$.

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