Show that $□$ PQRS is a rectangle if P ( $-$ 2,6), Q(3,1), R( $-$ 1, $-$ 3) and S( $-$ 6,2) are the co-ordinates of the vertices.

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Solution

$Given : P \equiv (- 2, 6), Q \equiv (3, 1), R \equiv (- 1, - 3), S \equiv (- 6, 2) Now, PQ = \sqrt{{[3 - (- 2)]}^{2} + {(1 - 6)}^{2}} = \sqrt{{(5)}^{2} + {(- 5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2} unit QR = \sqrt{{(- 1 - 3)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 4)}^{2}} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2} unit RS = \sqrt{{[- 6 - (- 1)]}^{2} + {[2 - (- 3)]}^{2}} = \sqrt{{(- 5)}^{2} + {(5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2} unit PS = \sqrt{{[- 6 - (- 2)]}^{2} + {(2 - 6)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 4)}^{2}} = \sqrt{16 + 16} = \sqrt{32} = 4 \sqrt{2} unit PR = \sqrt{{[- 1 - (- 2)]}^{2} + {(- 3 - 6)}^{2}} = \sqrt{{(1)}^{2} + {(- 9)}^{2}} = \sqrt{1 + 81} = \sqrt{82} unit QS = \sqrt{{(- 6 - 3)}^{2} + {(2 - 1)}^{2}} = \sqrt{{(- 9)}^{2} + {(1)}^{2}} = \sqrt{1 + 81} = \sqrt{82} unit$

$In □ PQRS, side PQ = side RS sideQR = side PS diagonal PR = diagonal QS ∴ □ PQRS is a rectangle .$

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Show that □PQRS is a rectangle if P (-2,6), Q(3,1), R(-1, -3) and S(-6,2) are the co-ordinates of the vertices.

Show that $□$ PQRS is a rectangle if P ( $-$ 2,6), Q(3,1), R( $-$ 1, $-$ 3) and S( $-$ 6,2) are the co-ordinates of the vertices.