CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


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

Solution

The correct option is A Collinear

As 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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image