Question

# The points $$({X}_{1},{Y}_{1})$$, $$({X}_{2},{Y}_{2})$$, $$({X}_{1},{Y}_{2})$$ and $$({X}_{2},{Y}_{1})$$ are always

A
Collinear
B
Concyclic
C
Vertices of a square
D
Vertices of rectangle

Solution

## The correct option is A CollinearAs all the points lie on the same line this implies that area of the triangle formed by the points is zero.So,using area formula in determinant form $${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })−{ X }_{ 2 }∗({ Y }_{ 1 }−{ Y }_{ 3 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$$${ X }_{ 1 }∗({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$Now,divide both sides by $${ X }_{ 1 }∗{ X }_{ 2 }∗{ X }_{ 3 }$$$$({ Y }_{ 2 }−{ Y }_{ 3 })+{ X }_{ 2 }∗({ Y }_{ 3 }−{ Y }_{ 1 })+{ X }_{ 3 }∗({ Y }_{ 1 }−{ Y }_{ 2 })=0$$$$\dfrac { { Y }_{ 2 }−{ Y }_{ 3 } }{ { X }_{ 2 }−{ X }_{ 3 } } +\dfrac { { Y }_{ 3 }−{ Y }_{ 1 } }{ { X }_{ 3 }−{ X }_{ 1 } } +\dfrac { { Y }_{ 1 }−{ Y }_{ 2 } }{ { X }_{ 1 }−{ X }_{ 2 } } =0$$Its collinear.Mathematics

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