Question

Prove that the points A(0,1), B(–2,3), C(6,7) and D(8,3) are the vertices of a rectangle ABCD.

Answer 1

$\: \red{\boxed{\sf{Solution.}}}$

$\tt \: We \: have$

$\: \: \: \: \: \: \: \: \: \: \tt \red{A } \implies(0, - 1)$

$\: \: \: \: \: \: \: \: \: \: \tt \red B \implies( - 2, 3)$

$\: \: \: \: \: \: \: \: \: \: \tt \red C \implies (6,7)$

$\rm and \: \: \: \: \: \: \: \: \: \tt \red D \implies(8,3)$

$\: \: \: \: \: \: \: \sf ∴ \red{AB} = \sqrt[]{( - 2 - 0) {}^{2} + (3 + 1) {}^{2} }$

$\: \: \: \: \: \sf = \sqrt{4 + 16 \: }$

$\: \: \: \: \: \sf = \sqrt{20 \: } = \sqrt{4 \times 5 \: }$

$\: \: \: \: \: \: \sf = 2 \sqrt{5 \: }$

$\: \: \: \: \: \sf \red{BC} = \sqrt{(6 + 2) {}^{2} + (7 - 3) {}^{2} } \:$

$\: \: \: \: \: \: \: \: \sf = \sqrt{64 + 16 \: }$

$\: \: \: \: \: \: \: \sf = \sqrt{80 \: } = \sqrt{16 \times 5 \: }$

$\: \: \: \: \: \: \: \: \: \sf 4 \sqrt{5 \: }$

$\: \: \: \: \sf \red{CD} = \sqrt{(8 - 6) {}^{2} + (3 - 7) {}^{2} }$

$\: \: \: \: = \sf \sqrt{4 + 16 \: } = \sqrt{20 \: }$

$\: \: \: \: \: \: = \sf \sqrt{4 \times 5 \: } = 2 \sqrt{5 \: }$

$\: \: \: \: \: \sf \red{DA} = \sqrt{(0 - 8) {}^{2} + ( - 1 - 3) {}^{2} }$

$\: \: \: \: \: \: \: = \sf \sqrt{64 + 16 \: }$

$\: \: \: \: \: \: = \sf \sqrt{80 \: } = \sqrt{16 \times 5 \: }$

$\: \: \: \: \: \: \: = \sf 4 \sqrt{5 \: }$

$\rm \purple{Here \: AB = CD \: and \: AD = BC}$

$\tt \: i.e., \: opposite \: sides \: of \\ \tt a \: quadrilateral \: are \: equal$

$\rm \pink{ ∴ ABCD \: is \: a \: parallelogram.}$

$\tt \: Now \: diagonal$

$\: \: \: \: \: \: \bf AC = \sqrt{(6 - 0) {}^{2} + (7 + 1) {}^{2} }$

$\: \: \: \: \: \: \: \sf \ = \sqrt{36 + 64 \: }$

$\: \: \: \: \: \: \: = \sf \sqrt{100 \: \: } = 10$

$\rm and \: diagonal \: \: \: \: BD \sf = \sqrt{(8 + 2) {}^{2} + (3 - 3) {}^{2} }$

$\: \: \: \: \: \: \sf = \sqrt{100 \: } = 10$

$\: \: \: \: \: \: \: \: \: \: \sf ∴ AC = BD$

$\sf \red{i.e., \: \: \: \: \: diagonals \: are \: equal}$

$\bf{ ∴ Points \: A, \: B, \: C, \: D \: are \: the \: vertices \: of \: a } \\ \bf{rectangle \: ABCD.}$

Answer 2

$\huge\textsf{\underline \red{AnsWer:}}$

➠ᴡᴇ ᴀʀᴇ ʜᴇʀᴇ ɢɪᴠᴇɴ ᴘᴏɪɴᴛs ᴀ,ʙ,ᴄ ᴀɴᴅ ᴅ ᴀɴx ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇsᴇ ᴀʀᴇ ᴛʜᴇ ᴠᴇʀᴛɪᴄs ᴏғ ʀᴇᴄᴛᴀɴɢʟᴇ ᴀʙᴄᴅ.

➠ᴘᴏɪɴᴛs :—

• ᴀ = (0,1)

• ʙ= (-2,3)

• ᴄ = (6,7)

• ᴅ = (8,3)

sᴏ, ʙᴇғᴏʀᴇ sᴏʟᴠɪɴɢ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,ɪᴛ ɪs ɴᴇᴄᴇssᴀʀʏ ᴛᴏ ᴋɴᴏᴡ ᴛʜᴇ ᴍᴇᴛʜᴏᴅ ᴛᴏ sᴏʟᴠᴇ ᴛʜɪs ǫᴜᴇsᴛɪᴏɴ ;

ɪғ ᴡᴇ ᴀʀᴇ ɢᴏɪɴɢ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇsᴇ ᴀʀᴇ ᴛʜᴇ ᴠᴇʀᴛɪᴄs ᴏғ ʀᴇᴄᴛᴀɴɢʟᴇ ᴛʜᴇɴ :-

• ᴏᴘᴘᴏsɪᴛᴇ sɪᴅᴇs ᴀʀᴇ ᴇǫᴜᴀʟ

• ᴅɪᴀɢɴᴏʟs ᴀʀᴇ ᴇǫᴜᴀʟ

ʟᴇᴛ's ʙᴇɢɪɴ ;

$\huge\textsf{\underline \red{Explanation:}}$

ғᴏʀᴍᴜʟᴀ, ᴛʜᴀᴛ ᴡᴇ ᴀʀᴇ ɢᴏɪɴɢ ᴛᴏ ᴜsᴇ :-

${\boxed{\sf \orange{Distance \: formula \: = \: \sqrt{ (x_2 - x_1)^2 +(y_2 - y_1)^2} }}}$

ɴᴏᴡ, ғɪɴᴅ ᴛʜᴇ ᴅɪsᴛᴀɴᴄᴇ ᴏғ sɪᴅᴇs;

$\setlength{\unitlength}{0.78cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ side\:1 }\put(8.1,5.3){ side\:2 }\put(11.5,7.5){ side\:3 }\put(8.1,9.5){ side\:4 }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(6,6){\line(5,3){5}}\end{picture}$

ɴᴏᴡ; [ʀᴇғᴇʀʀɪɴɢ ᴛᴏ ғɪɢᴜʀᴇ]

• ғɪʀsᴛ ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇ ᴏᴘᴘᴏsɪᴛᴇ sɪᴅᴇs ᴀʀᴇ ᴇǫᴜᴀʟ :-

$\star$$\normalsize\sf \blue{ \: Distance \: of \: AB(side \: 1):-}$

➠ $\normalsize\sf{ AB \: = \: \sqrt{(-2-0)^2 + (3+1)^2} }$

➠ $\normalsize\sf{ AB \: = \: \sqrt{(2)^2 + (4)^2} }$

➠ $\normalsize\sf{ AB \: = \: \sqrt{ 4 + 16} \: = \: \sqrt{20} }$

➠ $\normalsize\sf{ AB \: = \: \sqrt{2 \times\ 2 \times\ 5}\: = \: 2\sqrt{5} \: units }$

$\star$ $\normalsize\sf \pink{\: Distance \: of \: BC(side \: 2):-}$

➠ $\normalsize\sf{ BC \: = \: \sqrt{(6+2)^2 + (7-3)^2} }$

➠ $\normalsize\sf{ BC \: = \: \sqrt{(8)^2 + (4)^2} }$

➠ $\normalsize\sf{ BC \: = \: \sqrt{ 64 + 16} \: = \: \sqrt{80} }$

➠ $\normalsize\sf{ BC \: = \: \sqrt{ 2 \times\ 4 \times\ 4 \times\ 5} \: = \: 4\sqrt{5} \: units }$

$\star$ $\normalsize\sf \green{ \: Distance \: of \: CD(side \: 3):-}$

➠ $\normalsize\sf{ CD \: = \: \sqrt{(8-6)^2 + (3-7)^2} }$

➠ $\normalsize\sf{ CD \: = \: \sqrt{(2)^2 + (-4)^2} }$

➠ $\normalsize\sf{ CD \: = \: \sqrt{ 4 + 16} \: = \: \sqrt{20} }$

➠ $\normalsize\sf{ CD \: = \: \sqrt{ 2 \times\ 2 \times\ 5} \: = \: 2\sqrt{5} \: units}$

$\star$ $\normalsize\sf \purple{\: Distance \: of \: AD(side \: 4):-}$

➠ $\normalsize\sf{ AD \: = \: \sqrt{(0-8)^2 +(-1-3)^2} }$

➠ $\normalsize\sf{ AD \: = \: \sqrt{(-8)^2 + (4)^2} }$

➠ $\normalsize\sf{ AD \: = \: \sqrt{64 + 16} \: = \: \sqrt{80} }$

➠ $\normalsize\sf{ AD \: = \: \sqrt{ 2 \times\ 4 \times\ 4 \times\ 5} \: = \: 4\sqrt{5} \: units }$

ʜᴇʀᴇ ᴡᴇ ᴄᴀɴ sᴇᴇ ᴛʜᴀᴛ ᴏᴘᴘᴏsɪᴛᴇ sɪᴅᴇs ᴀʀᴇ ᴇǫᴜᴀʟ [ ᴀʙ = ᴄᴅ, ʙᴄ = ᴀᴅ ]

•ɴᴏᴡ, ᴡᴇ ʜᴀᴠᴇ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ᴅɪᴀɢɴᴏʟs ᴀʀᴇ ᴇǫᴜᴀʟ :-

$\star$ $\normalsize\sf \pink{ \: Distance \: of \: AC (diagnol \: 1):-}$

➠ $\normalsize\sf{ AC \: = \: \sqrt{(6-0)^2 + (7+1)^2 } }$

➠ $\normalsize\sf{ AC \: = \: \sqrt{(6)^2 + (8)^2} }$

➠ $\normalsize\sf{ AC \: = \: \sqrt{ 36 +64 } \: = \: \sqrt{100} }$

➠ $\normalsize\sf{ AC \: = \: 10 \: units}$

$\star$ $\normalsize\sf \green{\: Distance \: of \: BD(diagnol \: 2):-}$

➠ $\normalsize\sf{ BD \: = \: \sqrt{(8+8)^2 +(3-3)^2} }$

➠ $\normalsize\sf{ BD \: = \: \sqrt{(10)^2 + (0)^2} }$

➠ $\normalsize\sf{BD \: = \: \sqrt{100 + 0} \: \sqrt{10} }$

➠$\normalsize\sf{BD \: = \: 10}$

➠${\boxed{\sf \orange{Hence \: Prove }}}$