Question

show that show that the points (1,7 ),(4,2) ,(-1,1), (-4,4) are vertices of a square​

Answer 1

CORRECT QUESTION

Show that the Points (1 , 7), (4, 2) ,(-1 , 1) and (-4, 4) are the vertices of a sqaure.

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AnswEr

Let the Given Vertices be A (1, 7), B(4, 2), C(-1, 1) & D(-4, 4).

$\bf{\large{\underline{\sf{\pink{Using \; Distance \; Formula}}}}}$

$:\implies\large\boxed{\sf{\blue{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}$

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$\dag$ Finding the Distance of AB :-

$:\implies\sf\sqrt{(4 - 1)^2 + (2 - 7)^2}$

$:\implies\sf\sqrt{9 + 25}$

$:\implies\large\boxed{\sf{\red{\sqrt{34}}}}$

$\dag$ Finding the Distance of BC :-

$:\implies\sf\sqrt{(-1 - 4)^2 + (-1 - 1)^2}$

$:\implies\sf\sqrt{25 + 9}$

$:\implies\large\boxed{\sf{\red{\sqrt{34}}}}$

$\dag$ Finding the Distance of CD :-

$:\implies\sf\sqrt{(-4 +1)^2 + ( 4 + )^2}$

$:\implies\sf\sqrt{9 + 25}$

$:\implies\large\boxed{\sf{\red{\sqrt{34}}}}$

$\dag$ Finding the Distance of DA :-

$:\implies\sf\sqrt{(1 + 4)^2 + (7 - 4)^2}$

$:\implies\sf\sqrt{25 + 9}$

$:\implies\large\boxed{\sf{\red{\sqrt{34}}}}$

$\dag$ Finding the Distance of AC :-

$:\implies\sf\sqrt{(-1 -1)^2 + ( -1 - 7)^2}$

$:\implies\sf\sqrt{4 + 64}$

$:\implies\large\boxed{\sf{\pink{\sqrt{68}}}}$

$\dag$ Finding the Distance of BD :-

$:\implies\sf\sqrt{(-4 - 4)^2 + (4 - 2)^2}$

$:\implies\sf\sqrt{64 + 4}$

$:\implies\large\boxed{\sf{\pink{\sqrt{68}}}}$

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Here, We can see that

AC = BC = CD = DA & AC = BD

So, the Given points are the Vertices of a square.

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