Question

Show that the points (0,-1) ,(2,1) , (0,3) and (-2,1) are vertices of square.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Co-ordinate \: of \: a = (0, - 1) \\ \\ \tt: \implies Co-ordinate \: of \: b= (2, 1) \\ \\ \tt: \implies Co-ordinate \: of \: c = (0,3) \\ \\ \tt: \implies Co-ordinate \: of \: d = ( - 2, 1) \\ \\ \red{\underline \bold{To \: Show:}} \\ \tt: \implies Vertices \: are \: coordinate \: of \: square$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies AB = BC = CD= DA \: \: \: \: (Property \: of \: square) \\ \\ \bold{For \: AB \: Distance: } \\ \tt: \implies AB = \sqrt{ (x_{2} - x_{1})^{2} + ( y_{2} - y_{1})^{2} } \\ \\ \tt: \implies AB= \sqrt{(2 - 0)^{2} + ( - 1 - 1)^{2} } \\ \\ \tt: \implies AB = \sqrt{4 + 4 } \\ \\ \green{\tt: \implies AB = 2 \sqrt{2} \: units} \\ \\ \bold{Similarly \: for \: bc \: distance : } \\ \tt: \implies BC = \sqrt{(2 - 0)^{2} + {(1 - 3)}^{2} }$

$\tt:\implies BC = \sqrt{4 + 4} \\ \\ \tt: \implies BC = 2\sqrt{2} \: units \\ \\ \bold{For \: CD \:Distance : } \\ \tt: \implies CD = \sqrt{(0 - ( - 2))^{2} + ( 3 - 1)^{2} } \\ \\ \tt: \implies CD = \sqrt{4 + 4} \\ \\ \green{\tt: \implies CD = 2 \sqrt{2} \: units} \\ \\ \bold{For \:DA \: Distance : } \\ \tt: \implies DA = \sqrt{ (- 2 - 0)^{2} + (1 - ( - 1))^{2} } \\ \\ \tt: \implies DA = \sqrt{4 + 4} \\ \\ \green{\tt: \implies DA = 2 \sqrt{2} \: units} \\ \\ \green{\tt \therefore AB = BC = CD =DA}$