Question

Prove that the given points are vertices of a Square.

$\sf (0,5),(-1,2),(-6,0),(-3,6)$

Note:-

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Answer 1

⠀ || ☆.Solution.☆ ||

Given:-

• Vertices of Square , (0,5) , (-1,2) , (-6,0) , (-3,4)

prove :-

• These vertices are a square

⠀ || ☆.Explanation.☆ ||

We Know,

• Square have equal all side .

So, we prove all side are equal are not, if they are equal then we can say these vertices are of a square ,

if they are not equal they we can say these are not vertices are of a square .

_____________________

Let ABCD be a Square,

Where

• A(0,5) , B(-1,2) , C(-6,0) , D(-3,4)

Using Formula,

★Distance between two vertices = √[(x¹ - x)² - (y¹ - y)²]

Where ,

• (x,y) & (x¹ , y¹) are vertices.

_________________________

First Calculate AB,

➩ AB = √[(-1-0)² + (2-5)²]

➩ AB = √[(-1)² + (-3)²]

➩ AB = √[1 + 9]

➩ AB = √10.

_________________________

Now, Calculate BC,

➩ BC = √[(-6 + 1)² + (0-2)²]

➩ BC = √[(-5)² + (-2)²]

➩ BC = √(25 + 4)

➩ BC = √29

_________________________

Now, Calculate CD,

➩ CD = √[(-3 + 6)² + (4-0)²]

➩ CD = √[3² + 4²]

➩ CD = √(9 + 19)

➩ CD = √28

__________________,________

Now, Calculate DA ,

➩ DA = √[(-3-0)² + (4-5)²]

➩ DA = √[(-3)² + (-1)²]

➩ DA = √(9+1)

➩ DA = √10

_________________________

Here,

Distance of all side are unequal .

So, we can say That these are not vertices of a square.

That's Proved.

____________________

Answer 2

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