Question

$<h2 > Question$

Name the type of quadrilateral formed, if any, by the following points and give reason for your answer:
1. (–1,–2), (1,0) , (–1,2),
(–3,0)​

Answer 1

$\huge \sf{\underline{\underline {\boxed{\red{Correct \: Question.}}}}}$

$\rm \: Name \: the \: type \: of \: quadrilateral \:formed, \: if \: any ,\: by \: the \\ \rm following \: points, \: and \: given \: reasons \: for \: your \\ \rm answer.$

$\rm i) \: ( - 1, - 2)(1,0)( - 1,2)( - 3,0)$

$\huge{\red{\boxed{\overline{\underline{ \mid \tt{ \red {Solution \red {\mid}}}}}}}}$

$\rm( - 1, - 2)(1,0) (- 1,2)( - 3,0)$

$\rm \: Let \: A → \: ( - 1, - 2), \: B→ \: (1,0), \: C→ \: ( - 1,2) \: and \: D→ \: ( - 3,0)$

$\rm \: Then,$

$\rm \: \: \: \: \: \red{ side \: AB }= \sqrt{ [1 - ( - 1)] {}^{2} + ( 0- ( - 2) ] {}^{2} }$

$\implies \: \: \rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sqrt{(x_2 - x_1) {}^{2} + (y_2 - y_1) {}^{2} }$

$\rm = \: \: \: \: \sqrt{(2) {}^{2} + (2) {}^{2} } = \sqrt{4 + 4 \: }$

$\: \: \: \: \: \: \: \: \: \: \rm = \sqrt{8 \: } = \: \sqrt{4 \times 2 \: } = \: 2 \sqrt{2 \: }$

$\: \: \: \rm \: \: \: \red{ side \: BC }= \sqrt{( - 1 - 1) {}^{2} + (2 - 0) {}^{2} }$

$\: \: \: \: \: \: \rm = \: \: \sqrt{( - 2) {}^{2} + (2) {}^{2} } = \sqrt{4 + 4 \: }$

$\: \: \: \: \: \: \rm = \: \: \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \sqrt{2} = 2 \sqrt{2}$

$\rm \: \: \: \: \: \: \: \: \red{side \: CD} = \sqrt{[ - 3( - 1)] {}^{2} +( 0 - 2) {}^{2}}$

$\: \: \: \: \: \: = \: \rm \: \sqrt{( - 2) {}^{2} + ( - 2) {}^{2} } = \sqrt{4 + 4 \: }$

$\: \: \: \: \rm = \: \: \: \: \: \: \: \: \: \sqrt{8} = 2 \sqrt{2 \: }$

$\: \: \: \: \: \: \: \: \: \rm \red{side \: DA} \: = \sqrt{[ - 1 - ( - 3)] {}^{2} + ( - 2 - 0) {}^{2}}$

$\: \: \: \: \: \rm = \: \: \: \sqrt{(2) {}^{2} + ( - 2) {}^{2} } = \sqrt{4 + 4 \: }$

$\: \: \: \: \: \: \: \: \rm = \sqrt{8 \: } = \: \: 2 \sqrt{2 \: }$

$\rm \: We \: see \: that \: \red{AB = BC = CD = DA}$

$\rm \: i.e., \: \: \: \: \: \: \: \: \: \: \: all \: the \: four \: sides \: are \: equal \:$

$\therefore \: \: \: \: \: \: \rm The \: figure \: is \: either \: a \: square \: or \: a \: rhombus.$

$\rm \: Diagonal \: \: \: \: \: AC = \: \sqrt{[ - 1 - ( - 1) ] {}^{2} + [2 - ( - 2)] {}^{2} }$

$\: \: \: \: \: \: \rm = \: \: \: \: \sqrt{0 + 16 \: } = 4$

$\rm \: Diagonal \: \: \: \: BD = \: \sqrt{( - 3 - 1) {}^{2} + (0 - 0) {}^{2} }$

$\: \: \: \: \: \rm \: = \: \: \sqrt{16 + 0 \: } = \: 4$

$\rm \: Again \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf\red{AC = BD}$

$\bf \purple{(i.e., \: diagonals \: of \: the \: quadrilateral \: ABCD \: are \: equal.)}$

$\rm \: Therefore, \: \sf \red{ABCD} \rm \: is \: a \: \sf \red{square}.$

Answer 2

★$\huge\underline\bold\orange{Given}$★

$\bold{Measures \: of \: triangle \: (-1,-2),\: (1,0),}$

$\bold{(-1,3), \: (-3,0)}$

★$\huge\underline\bold\orange{Find\:out}$★

$\textbf{Name the type of quadrilateral and }$$\textbf{give reason for your answer}$

★$\huge\underline\bold\orange{Formula \: used}$★

$\textbf{Distance \: formula}$

$\sqrt{(x2-x1)^2+(y2-y1)^2}$

★$\huge\underline\bold\orange{Solution}$★

$\large\bold\blue{Distance \: of \: AB}$

$\bold{AB \: =}$$\sqrt{(x2-x1)^2+(y2-y1)^2}$

$\bold{AB \: =}$$\sqrt{(1-(-1))^2+(0-(-2))^2}$

$\bold{AB \: =}$$\sqrt{(1+1)^2+(0+2)^2}$

$\bold{AB \: =}$$\sqrt{(2)^2+(2)^2}$

$\bold{AB \: =}$$\sqrt{4+4}$

$\bold{AB \: =}$$\sqrt{8}$

$\bold{AB \: =}$${2}$$\sqrt{2}$unit

$\large\bold\blue{Distance \: of \: BC}$

$\bold{BC \: =}$$\sqrt{(-1-1)^2+(2-0)^2}$

$\bold{BC \: =}$$\sqrt{(-2)^2+(2)^2}$

$\bold{BC \: =}$$\sqrt{4+4}$

$\bold{BC \: =}$$\sqrt{8}$

$\bold{BC \: =}$${2}$$\sqrt{2}$unit

$\large\bold\blue{Distance \: of \: CD}$

$\bold{CD \: =}$$\sqrt{(-3-(-1))^2+(0-2)^2}$

$\bold{CD \: =}$$\sqrt{(-3+1)^2+(-2)^2}$

$\bold{CD \: =}$$\sqrt{(-2)^2+4}$

$\bold{CD \: =}$$\sqrt{4+4}$

$\bold{CD \: =}$$\sqrt{8}$

$\bold{CD \: =}$${2}$$\sqrt{2}$unit

$\large\bold\blue{Distance \: of \: DA}$

$\bold{DA \: =}$$\sqrt{(-1-(-3))^2+(-2-0)^2}$

$\bold{DA \: =}$$\sqrt{(-1+3)^2+(-2)^2}$

$\bold{DA \: =}$$\sqrt{(-2)^2+(-2)^2}$

$\bold{DA \: =}$$\sqrt{4+4}$

$\bold{DA \: =}$$\sqrt{8}$

$\bold{DA \: =}$${2}$$\sqrt{2}$unit

$\large\bold\blue{Distance \: of \: AC}$

$\bold{AC \: =}$$\sqrt{(-1-(-1))^2+(-2-2)^2}$

$\bold{AC \: =}$$\sqrt{(0)^2+(-4)^2}$

$\bold{AC \: =}$$\sqrt{16}$

$\bold{AC \: =}$${4}$unit

$\large\bold\blue{Distance \: of \: BD}$

$\bold{BD \: =}$$\sqrt{(-3-1)^2+(0-0)^2}$

$\bold{BD \: =}$$\sqrt{(-4)^2+(0)^2}$

$\bold{BD \: =}$$\sqrt{16}$

$\bold{BD \: = \: 4}$unit

$\textbf\red{Hence, AB=BC=CD=DA}$

$\textbf\red{AC=BD}$

.•.$\textbf\red{ABCD \: is \: a \: square}$

★$\textbf\red{All \: sides \: are \: equal \: in \: square}$★

★$\textbf\red{Diagonals \: are \: equal \: in \: square}$★