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Question Number 64311 by aseer imad last updated on 16/Jul/19
Commented by Tony Lin last updated on 17/Jul/19
letA′=A−(1,3)=(0,0),C′=C−(1,3)=(4,5) D′= [((((√2)/2) 0)),((0 ((√2)/2))) ] [((cos(π/4) −sin(π/4))),((sin(π/4) cos(π/4))) ] [(4),(5) ] = [(((1/2) −(1/2))),(((1/2) (1/2))) ] [(4),(5) ] = [((−(1/2))),(( (9/2))) ] ⇒D=D′+(1,3)=((1/2),((15)/2)) ∵AB^⇀ =DC^⇀ ∴B=(1,3)+((9/2),(1/2))=(((11)/2),(7/2))
$${letA}'={A}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{0},\mathrm{0}\right),{C}'={C}−\left(\mathrm{1},\mathrm{3}\right)=\left(\mathrm{4},\mathrm{5}\right) \\ $$$${D}'=\begin{bmatrix}{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{{cos}\frac{\pi}{\mathrm{4}}\:\:−{sin}\frac{\pi}{\mathrm{4}}}\\{{sin}\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:{cos}\frac{\pi}{\mathrm{4}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}}\\{\mathrm{5}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:=\begin{bmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\:\:\frac{\mathrm{9}}{\mathrm{2}}}\end{bmatrix} \\ $$$$\Rightarrow{D}={D}'+\left(\mathrm{1},\mathrm{3}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$$$\because{A}\overset{\rightharpoonup} {{B}}={D}\overset{\rightharpoonup} {{C}} \\ $$$$\therefore{B}=\left(\mathrm{1},\mathrm{3}\right)+\left(\frac{\mathrm{9}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\frac{\mathrm{11}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$
Answered by Tanmay chaudhury last updated on 16/Jul/19
AC (y−3)=((8−3)/(5−1))(x−1) y−3=(5/4)(x−1) 4y−5x−12+5=0 4y−5x−7=0 is ean AC let eqn AB (y−3)=m(x−1) angle betwee AB and AC is (π/4) tan(π/4)=((m−(5/4))/(1+m×(5/4))) 1+((5m)/4)=m−(5/4) ((5m)/4)−m=((−5)/4)−1 (m/4)=((−9)/4) m=−9 AB (y−3)=((−9)/1)(x−1) y−3+9x−9=0 y+9x−12=0 ←eqn AB DC (y−8)=((−9)/1)(x−5) y−8+9x−45=0 y+9x−53=0 ←eqn DC again tan(π/4)=(((5/4)−m)/(1+((5m)/4))) 1+((5m)/4)=(5/4)−m ((9m)/4)=(1/4) m=(1/9) eqn AD (y−3)=(1/9)(x−1) 9y−27−x+1=0 9y−x−26=0 ←eqn AD BC (y−8)=(1/9)(x−5) 9y−72=x−5 9y−x−67=0 ←eqn BC solve eqn DC y+9x−53=0 and AD 9y−x−26=0 to find D y+9(9y−26)−53=0 82y−234−53=0 y=((287)/(82))=(7/2) so x=9×(7/2)−26 D(((63−52)/2),(7/2)) so D(((11)/2),(7/2)) solveAB y+9x−12=0 BC 9y−x−67=0 y+9(9y−67)−12=0 82y=603+12 y=((615)/(82))=((15)/2) x=9×((15)/2)−67 x=((135−134)/2)=(1/2) B((1/2),((15)/2)) so A(1,3) B((1/2),((15)/2)),C(5,8) D(((11)/2),(7/2)) pls check Tanmay 16.07.19
$${AC}\:\:\left({y}−\mathrm{3}\right)=\frac{\mathrm{8}−\mathrm{3}}{\mathrm{5}−\mathrm{1}}\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{3}=\frac{\mathrm{5}}{\mathrm{4}}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{4}{y}−\mathrm{5}{x}−\mathrm{12}+\mathrm{5}=\mathrm{0}\:\:\:\:\mathrm{4}{y}−\mathrm{5}{x}−\mathrm{7}=\mathrm{0}\:\:{is}\:{ean}\:{AC} \\ $$$${let}\:{eqn}\:{AB}\:\left({y}−\mathrm{3}\right)={m}\left({x}−\mathrm{1}\right) \\ $$$${angle}\:{betwee}\:{AB}\:{and}\:{AC}\:{is}\:\frac{\pi}{\mathrm{4}} \\ $$$${tan}\frac{\pi}{\mathrm{4}}=\frac{{m}−\frac{\mathrm{5}}{\mathrm{4}}}{\mathrm{1}+{m}×\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{1}+\frac{\mathrm{5}{m}}{\mathrm{4}}={m}−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\frac{\mathrm{5}{m}}{\mathrm{4}}−{m}=\frac{−\mathrm{5}}{\mathrm{4}}−\mathrm{1} \\ $$$$\frac{{m}}{\mathrm{4}}=\frac{−\mathrm{9}}{\mathrm{4}}\:\:\:{m}=−\mathrm{9} \\ $$$${AB}\:\:\left({y}−\mathrm{3}\right)=\frac{−\mathrm{9}}{\mathrm{1}}\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{3}+\mathrm{9}{x}−\mathrm{9}=\mathrm{0} \\ $$$${y}+\mathrm{9}{x}−\mathrm{12}=\mathrm{0}\:\leftarrow{eqn}\:{AB} \\ $$$${DC}\:\:\left({y}−\mathrm{8}\right)=\frac{−\mathrm{9}}{\mathrm{1}}\left({x}−\mathrm{5}\right) \\ $$$${y}−\mathrm{8}+\mathrm{9}{x}−\mathrm{45}=\mathrm{0} \\ $$$${y}+\mathrm{9}{x}−\mathrm{53}=\mathrm{0}\:\:\leftarrow{eqn}\:{DC} \\ $$$${again}\:{tan}\frac{\pi}{\mathrm{4}}=\frac{\frac{\mathrm{5}}{\mathrm{4}}−{m}}{\mathrm{1}+\frac{\mathrm{5}{m}}{\mathrm{4}}} \\ $$$$\mathrm{1}+\frac{\mathrm{5}{m}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{4}}−{m} \\ $$$$\frac{\mathrm{9}{m}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\:\:{m}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${eqn}\:\:{AD}\:\:\left({y}−\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{9}{y}−\mathrm{27}−{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{9}{y}−{x}−\mathrm{26}=\mathrm{0}\:\:\leftarrow{eqn}\:{AD}\: \\ $$$${BC}\:\left({y}−\mathrm{8}\right)=\frac{\mathrm{1}}{\mathrm{9}}\left({x}−\mathrm{5}\right) \\ $$$$\mathrm{9}{y}−\mathrm{72}={x}−\mathrm{5} \\ $$$$\mathrm{9}{y}−{x}−\mathrm{67}=\mathrm{0}\:\leftarrow{eqn}\:{BC} \\ $$$${solve}\:{eqn}\:{DC}\:\:{y}+\mathrm{9}{x}−\mathrm{53}=\mathrm{0}\:{and}\:{AD}\:\mathrm{9}{y}−{x}−\mathrm{26}=\mathrm{0} \\ $$$${to}\:{find}\:{D}\:\:{y}+\mathrm{9}\left(\mathrm{9}{y}−\mathrm{26}\right)−\mathrm{53}=\mathrm{0} \\ $$$$\mathrm{82}{y}−\mathrm{234}−\mathrm{53}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{287}}{\mathrm{82}}=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:{so}\:{x}=\mathrm{9}×\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{26} \\ $$$${D}\left(\frac{\mathrm{63}−\mathrm{52}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right)\:\:{so}\:\boldsymbol{{D}}\left(\frac{\mathrm{11}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$$${solveAB}\:{y}+\mathrm{9}{x}−\mathrm{12}=\mathrm{0}\:{BC}\:\mathrm{9}{y}−{x}−\mathrm{67}=\mathrm{0} \\ $$$${y}+\mathrm{9}\left(\mathrm{9}{y}−\mathrm{67}\right)−\mathrm{12}=\mathrm{0} \\ $$$$\mathrm{82}{y}=\mathrm{603}+\mathrm{12}\:\:\:\:\:{y}=\frac{\mathrm{615}}{\mathrm{82}}=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$${x}=\mathrm{9}×\frac{\mathrm{15}}{\mathrm{2}}−\mathrm{67} \\ $$$${x}=\frac{\mathrm{135}−\mathrm{134}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${B}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right)\:{so} \\ $$$${A}\left(\mathrm{1},\mathrm{3}\right)\:{B}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right),{C}\left(\mathrm{5},\mathrm{8}\right)\:\:{D}\left(\frac{\mathrm{11}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right)\:\:{pls}\:{check}\: \\ $$$${Tanmay}\:\mathrm{16}.\mathrm{07}.\mathrm{19} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aseer imad last updated on 16/Jul/19
Thank you very much sir,it is right. Could've avoid the complexity of finding all equations. Instead finding one vertex and use mid point to find the other...
Commented by Tanmay chaudhury last updated on 17/Jul/19
i am going to rechek the steps..
$${i}\:{am}\:{going}\:{to}\:{rechek}\:{the}\:{steps}.. \\ $$
Answered by mr W last updated on 16/Jul/19
M=midpoint of AC=midpoint of BD M(((1+5)/2),((3+8)/2))=M(3,((11)/2)) Δx_(MB) =Δy_(AM) =((11)/2)−3=(5/2) ⇒x_B =3+(5/2)=((11)/2) Δy_(MB) =Δx_(AM) =3−1=2 ⇒y_B =((11)/2)−2=(7/2) ((x_D +((11)/2))/2)=3 ⇒x_D =6−((11)/2)=(1/2) ((y_D +(7/2))/2)=((11)/2) ⇒y_D =11−(7/2)=((15)/2) ⇒B(((11)/2),(7/2)), D((1/2),((15)/2))
$${M}={midpoint}\:{of}\:{AC}={midpoint}\:{of}\:{BD} \\ $$$${M}\left(\frac{\mathrm{1}+\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}+\mathrm{8}}{\mathrm{2}}\right)={M}\left(\mathrm{3},\frac{\mathrm{11}}{\mathrm{2}}\right) \\ $$$$\Delta{x}_{{MB}} =\Delta{y}_{{AM}} =\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{3}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{B}} =\mathrm{3}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\Delta{y}_{{MB}} =\Delta{x}_{{AM}} =\mathrm{3}−\mathrm{1}=\mathrm{2} \\ $$$$\Rightarrow{y}_{{B}} =\frac{\mathrm{11}}{\mathrm{2}}−\mathrm{2}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\frac{{x}_{{D}} +\frac{\mathrm{11}}{\mathrm{2}}}{\mathrm{2}}=\mathrm{3} \\ $$$$\Rightarrow{x}_{{D}} =\mathrm{6}−\frac{\mathrm{11}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{y}_{{D}} +\frac{\mathrm{7}}{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\Rightarrow{y}_{{D}} =\mathrm{11}−\frac{\mathrm{7}}{\mathrm{2}}=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$$\Rightarrow{B}\left(\frac{\mathrm{11}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right),\:{D}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right) \\ $$
Commented by mr W last updated on 17/Jul/19
Answered by ajfour last updated on 17/Jul/19
Commented by ajfour last updated on 17/Jul/19
B(1+4+(1/2) , 3+(1/2)) ⇒ B(((11)/2),(7/2)) D(1−(1/2), 3+(1/2)+4) ⇒ D((1/2),((15)/2)) .
$${B}\left(\mathrm{1}+\mathrm{4}+\frac{\mathrm{1}}{\mathrm{2}}\:,\:\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow\:{B}\left(\frac{\mathrm{11}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$$${D}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{4}\right)\:\Rightarrow\:{D}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{15}}{\mathrm{2}}\right)\:. \\ $$
Commented by aseer imad last updated on 22/Jul/19
Cool! thanks

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