∑i=14(xi2+yi2)≤2x1x3+2x2x4+2y2y3+2y1y4 the points (x1,y1),(x2,y2),(x3,y3),(x4,y4) are

Let A≡(x1,y1),B≡(x2,y2),C≡(x3,y3),D≡(x4,y4)Givenx12+x22+x32+x42+y12+y22+y32+y42−2x1x3−2x2x4−2y2y3−2y1y4≤0x1=x3,x2=x4,y2y3,y1=y4(or)x1+x22=x3+x42andy1+y22=y4+y32Hence , AB and CD bisect each other. Therefore ABCD is a parallelogramAlso,AB2=(x1−x2)2+(y1−y2)2=(x3−x4)2+(y4−y3)2=CD2 Thus, ABCD is a parallelogram and AB = CD. Hence, it is a rectangle.

∑i=14(xi2+yi2)≤2x1x3+2x2x4+2y2y3+2y1y4 the points (x1,y1),(x2,y2),(x3,y3),(x4,y4) are

Let A≡(x1,y1),B≡(x2,y2),C≡(x3,y3),D≡(x4,y4)Givenx12+x22+x32+x42+y12+y22+y32+y42−2x1x3−2x2x4−2y2y3−2y1y4≤0x1=x3,x2=x4,y2y3,y1=y4(or)x1+x22=x3+x42andy1+y22=y4+y32Hence , AB and CD bisect each other. Therefore ABCD is a parallelogramAlso,AB2=(x1−x2)2+(y1−y2)2=(x3−x4)2+(y4−y3)2=CD2 Thus, ABCD is a parallelogram and AB = CD. Hence, it is a rectangle.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The vertices of a rectangle

Collinear

The vertices of a trapezium

Rhombus

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let $A \equiv (x_{1}, y_{1}), B \equiv (x_{2}, y_{2}), C \equiv (x_{3}, y_{3}), D \equiv (x_{4}, y_{4})$

$Given x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} - 2 x_{1} x_{3} - 2 x_{2} x_{4} - 2 y_{2} y_{3} - 2 y_{1} y_{4} \leq 0$

$x_{1} = x_{3}, x_{2} = x_{4}, y_{2} y_{3}, y_{1} = y_{4} (or) \frac{x_{1} + x_{2}}{2} = \frac{x_{3} + x_{4}}{2} and \frac{y_{1} + y_{2}}{2} = \frac{y_{4} + y_{3}}{2}$