Show that $□$ LMNP is a square if L( $-$ 2,4) M(4,2), N(2, $-$ 4) and P ( $-$ 4, $-$ 2) are the co-ordinate of the vertices.

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Solution

$Given : L \equiv (- 2, 4), M \equiv (4, 2), N \equiv (2, - 4), P \equiv (- 4, - 2) Now, LM = \sqrt{{[4 - (- 2)]}^{2} + {(2 - 4)}^{2}} = \sqrt{{(6)}^{2} + {(- 2)}^{2}} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10} units MN = \sqrt{{(2 - 4)}^{2} + {(- 4 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 6)}^{2}} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} units NP = \sqrt{{(- 4 - 2)}^{2} + {[- 2 - (- 4)]}^{2}} = \sqrt{{(- 6)}^{2} + {(2)}^{2}} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10} units LP = \sqrt{{[- 4 - (- 2)]}^{2} + {(- 2 - 4)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 6)}^{2}} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} units LN = \sqrt{{[2 - (- 2)]}^{2} + {(- 4 - 4)}^{2}} = \sqrt{{(4)}^{2} + {(- 8)}^{2}} = \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5} units MP = \sqrt{{(- 4 - 4)}^{2} + {(- 2 - 2)}^{2}} = \sqrt{{(- 8)}^{2} + {(- 4)}^{2}} = \sqrt{16 + 64} = \sqrt{80} = 4 \sqrt{5} units$

$In □ PQRS, side LM = side MN = side NP = side LP and diagonal LN = diagonal MP . ∴ □ PQRS is a square .$

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