If `sum_(i=1)^4(x1^2+y 1^2)lt=2x_1x_3+2x_2x_4+2y_2y_3+2y_1y_4,` the points `(x_1, y_1),(x_2,y_2),(x_3, y_3),(x_4,y_4)` are

Options

A. the vertices of a rectangle
B. collinear
C. the vertices of a trapezium
D. none of these

Answer

A

Explanation

Let `A-=(x_1,y_1),B-=(x_2,y_2), C-=(x_3,y_3), D-=(x_4,y_4)`
Given
`x_1^2+x_2^(2)+x_4^(2)+y_1^(2)+y_2^(2)+y_3^(2) +y_4^2-2x_1x_3-2x_(2)x_4-2y_(2)y_(3)-2y_(1)y_4le0`
or `(x_1-x_3)^2+(x_2-x_4)^2+(y_2-y_3)^2+(y_2-y_3)^2+(y_1-y_4)^2le0`
or `(x_1+x_2)/(2)=(x_3+x_4)/(2) and (y_1+y_2)/(2)=(y_4+y_3)/(2)`
Hence AB and CD bisect each other. therefore, ACBD is a parallelogram. Also,
`AB^2=(x_1-x_2)^2+(y_1-y_2)^2`
`=(x_3-x_4)^2+(y_4-y_3)^2`
`=CD^2`