Question

A(3,2),B(-2,-3), C(2,3), form a trian
gle or not?​

Answer 1

Answer:

Yes, they form a triangle.

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Step-by-step explanation:

In order for any three points to form a triangle, the sum of the length of two sides must be greater than the length of the third side.

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So, we'll need to find the length of AB, BC and CA, and then compare them. We'll find the lengths of the sides using the distance formula.

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For any two points (x₁, y₁) and (x₂, y₂), the distance between the two points is given by the Distance formula.

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

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AB's Distance:

A ➝ (3, 2)

B ➝ (-2, -3)

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$\sf \dashrightarrow AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow AB = \sqrt{(-2 - 3)^2 + (-3 - 2)^2}$

$\sf \dashrightarrow AB = \sqrt{(-5)^2 + (-5)^2}$

$\sf \dashrightarrow AB = \sqrt{25 + 25}$

$\sf \dashrightarrow AB = \sqrt{50}$

$\sf \dashrightarrow AB = 5\sqrt{2} \ units.$

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BC's Distance:

B ➝ (-2, -3)

C ➝ (2, 3)

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$\sf \dashrightarrow BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow BC = \sqrt{(2 - (-2))^2 + (3 - (-3))^2}$

$\sf \dashrightarrow BC = \sqrt{(2 + 2)^2 + (3 + 3)^2}$

$\sf \dashrightarrow BC = \sqrt{(4)^2 + (6)^2}$

$\sf \dashrightarrow BC = \sqrt{16 + 36}$

$\sf \dashrightarrow BC = \sqrt{52}$

$\sf \dashrightarrow BC = 2\sqrt{13} \ units.$

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CA's Distance:

C ➝ (2, 3)

A ➝ (3, 2)

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$\sf \dashrightarrow CA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf \dashrightarrow CA = \sqrt{(3 - 2)^2 + (2 - 3)^2}$

$\sf \dashrightarrow CA = \sqrt{(1)^2 + (-1)^2}$

$\sf \dashrightarrow CA = \sqrt{1 + 1}$

$\sf \dashrightarrow CA = \sqrt{2} \ units.$

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Now;

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Comparing AB + BC with CA we get;

$\dashrightarrow \sf \ AB + BC > CA$

$\dashrightarrow \sf \ 5\sqrt{2} + 2\sqrt{13} > \sqrt{2}$

$\dashrightarrow \sf \ 7.071 + 7.211 > 1.414 \ \big(Approximate \ values\big)$

$\dashrightarrow \sf \ 14.282 > 1.414$

AB + BC is greater than CA.

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Comparing BC + CA with AB we get;

$\dashrightarrow \sf \ BC + CA > AB$

$\dashrightarrow \sf \ 2\sqrt{13} + \sqrt{2} > 5\sqrt{2}$

$\dashrightarrow \sf \ 7.211 + 1.414 > 7.071 \ \big(Approximate \ values\big)$

$\dashrightarrow \sf \ 8.625 > 1.414$

BC + CA is greater than AB.

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Comparing CA + AB with BC we get;

$\dashrightarrow \sf \ CA + AB > BC$

$\dashrightarrow \sf \ \sqrt{2} + 5\sqrt{2} \ > \ 2\sqrt{13}$

$\dashrightarrow \sf \ \sqrt{2}(1 + 5) \ > \ 2\sqrt{13}$

$\dashrightarrow \sf \ 6\sqrt{2} \ > \ 2\sqrt{13}$

$\dashrightarrow \sf \ 8.485 \ > \ 7.211 \ \big(Approximate \ values\big)$

CA + AB is greater than BC.

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All three conditions are satisfied, therefore the given points form a triangle.

Answer 2

Answer:

$:\implies$Yes,they form a triangle.

hope it helps uh :)