Question

prove that the points (2 -2) (-2 1) and (5 2) are the vertices of a right angled triangle

Answer 1

Let us name the points (2, -2) (-2, 1) and (5. 2) A, B and C respectively.

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For any three points to form a right-angled triangle, the square of the length of the hypotenuse (longest side) must be equal to the sum of the squares of the other two sides of the triangle. [Pythagoras' Theorem]

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Since we'll need the length of all three sides, we'll use the Distance formula to calculate the lengths.

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For any two points with the co-ordinates (x₁, y₁) and (x₂, y₂), the formula used to calculate the distance is;

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

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Length of AB;

A = (2, -2)

B = (-2, 1)

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

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

$\dashrightarrow \sf AB = \sqrt{(-4)^{2} + (1 + 2)^{2}}$

$\dashrightarrow \sf AB = \sqrt{(-4)^{2} + (3)^{2}}$

$\dashrightarrow \sf AB = \sqrt{16 + 9}$

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

$\dashrightarrow \sf AB = \pm \ 5$

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Distance can't be negative, therefore AB = 5 units.

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Length of BC;

B = (-2, 1)

C = (5, 2)

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

$\dashrightarrow \sf BC = \sqrt{(5 - (-2))^{2} + (2 - 1)^{2}}$

$\dashrightarrow \sf BC = \sqrt{(5+ 2)^{2} + (1)^{2}}$

$\dashrightarrow \sf BC = \sqrt{(7)^{2} + 1}$

$\dashrightarrow \sf BC = \sqrt{49 + 1}$

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

$\dashrightarrow \sf BC = \sqrt{2 \times 5 \times 5}$

$\dashrightarrow \sf BC = 5\sqrt{2}$

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Therefore, BC = 5√2 units.

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Length of CA;

C = (5, 2)

A = (2, -2)

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

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

$\dashrightarrow \sf CA = \sqrt{(-3)^{2} + (-4)^{2}}$

$\dashrightarrow \sf CA = \sqrt{9 + 16}$

$\dashrightarrow \sf CA = \sqrt{25}$

$\dashrightarrow \sf CA = \pm \ 5$

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Distance can't be negative, therefore, CA = 5 units.

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On comparing the values of all three sides, BC is the longest, therefore by applying Pythagoras' theorem we get;

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⇒ BC² = AB² + CA²

⇒ [5√2]² = [5]² + [5]²

⇒ 25 × 2 = 25 + 25

⇒ 50 = 50

⇒ LHS = RHS

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Therefore, the three given points form a right-angled triangle. Hence proved.