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If $\sum_{i = 1}^{4} (x_{i}^{2} + y_{i}^{2}) \leq 2 x_{1} x_{3} + 2 x_{2} x_{4} + 2 y_{2} y_{3} + 2 y_{1} y_{4},$ the points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3} y_{3}), (x_{4}, y_{4})$ are

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a

the vertices of a rectangle

b

collinear

c

the vertices of a trapezium

d

none of these

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detailed solution

Correct option is A

Let A≡x1,y1, B≡x2,y2, C≡x3,y3, D≡x4,y4given x12+x22+x32+x42+y12+y22+y32+y42−2x1x3−2x2x4−2y2y3−2y1y4≤0or x1−x32+x2−x42+y2−y32+y1−y42≤0or x1=x3,x2=x4, y2=y3, y1=y4or x1+x22=x3+x42 andy1+y22=y4+y32Hence, AB and CD bisect each other,Therefore, ACBD is a parallelogram. Also,AB2=x1−x22+y1−y22=x3−x42+y4−y32=CD2Thus, ACBD is a parallelogram and AB=CD.Hence, It is a rectangle.

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Let $A (0, 1), B (1, 1), C (1, - 1), D (- 1, 0)$ be four points. If P is any other point, then $PA + PB + PC + PD \geq d,$ when $[d]$ is ___, where [.] represents greatest integer.

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