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The-points-A-B-C-represent-the-complex-numbers-z-1-z-2-z-3-respectively-And-G-is-the-centroid-of-the-triangle-A-B-C-if-4z-1-z-2-z-3-0-show-that-the-origin-is-the-mid-point-of

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Question Number 51320 by Tawa1 last updated on 25/Dec/18
The points A, B, C represent the complex numbers z_1 , z_2 , z_3 respectively. And G is the centroid of the triangle A B C, if 4z_1 + z_2 + z_3 = 0, show that the origin is the mid point of AG.
$$\mathrm{The}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{represent}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:\:\mathrm{z}_{\mathrm{1}} ,\:\mathrm{z}_{\mathrm{2}} ,\:\mathrm{z}_{\mathrm{3}} \: \\ $$$$\mathrm{respectively}.\:\mathrm{And}\:\mathrm{G}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{A}\:\mathrm{B}\:\mathrm{C},\:\:\mathrm{if} \\ $$$$\mathrm{4z}_{\mathrm{1}} \:+\:\mathrm{z}_{\mathrm{2}} \:+\:\mathrm{z}_{\mathrm{3}} \:\:=\:\:\mathrm{0},\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}\:\mathrm{point}\:\mathrm{of}\:\:\mathrm{AG}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18
A(x_1 ,y_1 ) B(x_2 ,y_2 ) C(x_3 ,y_3 ) G(((x_1 +x_2 +x_3 )/3),((y_1 +y_2 +y_3 )/3)) mid point AG{(1/2)(x_1 +((x_1 +x_2 +x_3 )/3)),(1/2)(y_1 +((y_1 +y_2 +y_3 )/3))} ={(1/2)(((4x_1 +x_2 +x_3 )/3)),(1/2)(((4y_1 +y_2 +y_3 )/3))}→(0,0) already given) hence proved...
$${A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)\:{C}\left({x}_{\mathrm{3}} ,{y}_{\mathrm{3}} \right)\:{G}\left(\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }{\mathrm{3}},\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{y}_{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$${mid}\:{point}\:{AG}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left({x}_{\mathrm{1}} +\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }{\mathrm{3}}\right),\frac{\mathrm{1}}{\mathrm{2}}\left({y}_{\mathrm{1}} +\frac{{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{y}_{\mathrm{3}} }{\mathrm{3}}\right)\right\} \\ $$$$\left.=\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }{\mathrm{3}}\right),\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{4}{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{y}_{\mathrm{3}} }{\mathrm{3}}\right)\right\}\rightarrow\left(\mathrm{0},\mathrm{0}\right)\:{already}\:{given}\right) \\ $$$${hence}\:{proved}… \\ $$$$ \\ $$
Commented by Tawa1 last updated on 26/Dec/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 26/Dec/18
Sir but why is midpoint of AG not divided by 2, like (((x_1 + ((x_1 + x_2 + x_3 )/3))/2), .. i thought midpoint is divided by 2 sir. please let me understand sir
$$\mathrm{Sir}\:\mathrm{but}\:\mathrm{why}\:\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\:\mathrm{AG}\:\:\mathrm{not}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{2},\:\:\mathrm{like} \\ $$$$\left(\frac{\mathrm{x}_{\mathrm{1}} \:+\:\frac{\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{x}_{\mathrm{2}} \:+\:\mathrm{x}_{\mathrm{3}} }{\mathrm{3}}}{\mathrm{2}},\:..\right. \\ $$$$\mathrm{i}\:\mathrm{thought}\:\mathrm{midpoint}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{2}\:\mathrm{sir}.\:\:\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{understand}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18
yes you are right...let me[rectify...
$${yes}\:{you}\:{are}\:{right}…{let}\:{me}\left[{rectify}…\right. \\ $$
Commented by Tawa1 last updated on 26/Dec/18
God bless you sir. I really appreciate your time sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$

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