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A point P moves along the x-axis. Let L(P) = PA + PB where A = (0, 2)…
Question
A point P moves along the x-axis. Let L(P) = PA + PB where A = (0, 2) and B = (10, 3). Minimize L over all positions of P, then report the value of (minimum of L)^2.
✓ Verified answer: 125checked by our engine — not a guess
Step-by-step solution
A and B are on the same side of the x-axis (both above).
Reflect B = (10, 3) across the x-axis to B' = (10, -3).
For any P on the axis, PB = PB', so PA + PB = PA + PB' >= AB', with equality when P lies on segment AB'.
Thus the minimum of L equals AB' = sqrt((10-0)^2 + (-3-2)^2) = sqrt(100 + 25) = sqrt(125).
Therefore (min L)^2 = 125.
Final answer125
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