The points A and B are (1, 2) and (5, 4). The point P on the x-axis for which \(PA^{2} + PB^{2}\) is least is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (3, 0)
Let P = (x, 0). By the distance formula,
\( S(x) = PA^{2} + PB^{2} = (x - 1)^{2} + 4 + (x - 5)^{2} + 16 \)
Differentiating and setting the derivative equal to zero:
\( S'(x) = 2(x - 1) + 2(x - 5) = 4x - 12 \)
\( 4x - 12 = 0 \), so x = 3
Since \(S''(x) = 4 \gt 0\), this gives the least value:
\( S(3) = 4 + 4 + 4 + 16 = 28 \)
So P = (3, 0).
Why the other options are incorrect
Option (1) (2, 0) — Incorrect. S(2) = 1 + 4 + 9 + 16 = 30, more than 28.
Option (2) \(\left(\displaystyle\frac{7}{3}, 0\right)\) — Incorrect. This point makes the sum of the distances PA + PB least, not the sum of their squares; here \( S\left(\frac{7}{3}\right) = \frac{80}{9} + 20 \gt 28 \).
Option (4) (4, 0) — Incorrect. S(4) = 9 + 4 + 1 + 16 = 30, more than 28.