First slide

Cartesian plane

Question

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

Moderate

A

the vertices of a rectangle
B

collinear
C

the vertices of a trapezium
D

none of these

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
$x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2} - 2 x_{1} x_{3} - 2 x_{2} x_{4} - 2 y_{2} y_{3} - 2 y_{1} y_{4} \leq 0$
or ${(x_{1} - x_{3})}^{2} + {(x_{2} - x_{4})}^{2} + {(y_{2} - y_{3})}^{2} + {(y_{1} - y_{4})}^{2} \leq 0$
or $x_{1} = x_{3}, x_{2} = x_{4}, y_{2} = y_{3}, y_{1} = y_{4}$
or $\frac{x_{1} + x_{2}}{2} = \frac{x_{3} + x_{4}}{2}$ and $\frac{y_{1} + y_{2}}{2} = \frac{y_{4} + y_{3}}{2}$
Hence, $AB$ and $CD$ bisect each other,
Therefore, $ACBD$ is a parallelogram. Also,
${AB}^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$
$= {(x_{3} - x_{4})}^{2} + {(y_{4} - y_{3})}^{2}$
$= {CD}^{2}$
Thus, $ACBD$ is a parallelogram and $AB = CD .$ Hence, It is a rectangle.

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