Question

the points(1, 2), b(5, 4) c(3, 8) d(-1, 6) are the vertices of a quadrilateral using distance formula identify the type​

Answer 1

GivEn:

• A (1,2)
• B (5,4)
• C (3,8)
• D (-1,6)

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To find:

• Using distance Formula identify which type of quadrilateral is it?

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SoluTion:

$\bf Here = \begin{cases} & \text{A (1,2) } \\ & \text{B (5,4)} \\ & \text{C (3,8)} \\ & \text{D (-1,6)}\end{cases}$

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As we know that,

$\star\;{\boxed{\sf{\purple{Distance\;Formula = \sqrt{( x_2 - x_1 )^2 + ( y_2 - y_1 )^2}}}}}$

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${\underline{\sf{\bigstar\;Now,\;Using\;Distance\;Formula,\; distance\;of\;sides\;are\;:}}}$

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★ Distance AB :-

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$:\implies\sf \sqrt{( 5 - 1)^2 + (4 - 2)^2}$

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

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$:\implies\sf \sqrt{16 + 4}$

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$:\implies\sf \sqrt{20}$

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$:\implies\bf \pink{2 \sqrt{5}}$

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★ Distance BC :-

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$:\implies\sf \sqrt{(3 - 5)^2 + (8 - 4)^2}$

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

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$:\implies\sf \sqrt{4 + 16}$

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$:\implies\sf \sqrt{20}$

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$:\implies\bf \pink{2 \sqrt{5}}$

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★ Distance CD :-

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$:\implies\sf \sqrt{(-1 -3)^2 + (6 - 8)^2}$

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

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$:\implies\sf \sqrt{16 + 4}$

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$:\implies\bf \pink{2 \sqrt{5}}$

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★ Distance DA :-

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$:\implies\sf \sqrt{(1 + 1)^2 + (2 - 6)^2}$

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

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$:\implies\sf \sqrt{4 + 16}$

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$:\implies\bf \pink{2 \sqrt{5}}$

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$\therefore$ AB = BC = CD = DA

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All sides are equal.

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${\underline{\sf{\bigstar\;Now,\;Using\;Distance\;Formula,\; distance\;of\; diagonal\;are\;:}}}$

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★ Distance AC :-

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$:\implies\sf \sqrt{(3 - 1)^2 + (6 - 2)^2}$

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

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$:\implies\sf \sqrt{4 + 16}$

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$:\implies\bf \blue{2 \sqrt{5}}$

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★ Distance BD :-

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$:\implies\sf \sqrt{( - 1 + 5)^2 + (6 - 4)^2}$

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

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$:\implies\sf \sqrt{ 16 + 4}$

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$:\implies\bf \blue{2 \sqrt{5}}$

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$\therefore$ AC = BD

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Both Diagonal are equal.

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★ Hence, We can see that all sides and diagonal are of equal distance.

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$\therefore$ Point A, B, C and D are vertices of a square.