Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

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Solution

Let A (2, – 2), B (– 2, 1) and C (5, 2) be the vertices of ΔABC.

$A B = \sqrt{(2 - (- 2))^{2} + (- 2 - 1)^{2}} = \sqrt{4^{2} + (- 3)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

$B C = \sqrt{(- 2 - 5)^{2} + (1 - 2)^{2}} = \sqrt{(- 7)^{2} + (- 1)^{2}} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2}$

$C A = \sqrt{(5 - 2)^{2} + (2 - (- 2))^{2}} = \sqrt{(3)^{2} + (4)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$

$A B^{2} + C A^{2} = (5)^{2} + (5)^{2} = 25 + 25 = 50$

$B C^{2} = (5 \sqrt{2})^{2} = 50$

$A B^{2} + C A^{2} = B C^{2}$

In ΔABC,

$A B^{2} + C A^{2} = B C^{2}$

∴ ΔABC is a right triangle right angled at $∠ A$ . (Converse of Pythagoras theorem)

Length of hypotenuse, $B C = 5 \sqrt{2}$

Area of ΔABC

$= \frac{1}{2} \times A B \times C A = \frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12.5 s q . u n i t s$

$[S i n c e, A r e a o f t r i a n g l e = \frac{1}{2} \times b a s e \times h e i g h t]$

Hence required area 12.5 sq units

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