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Question Number 132765 by Ahmed1hamouda last updated on 16/Feb/21

Answered by mr W last updated on 16/Feb/21

Commented by mr W last updated on 16/Feb/21

G(((7a)/4),((5a)/4))

$${G}\left(\frac{\mathrm{7}{a}}{\mathrm{4}},\frac{\mathrm{5}{a}}{\mathrm{4}}\right) \\ $$

Answered by TheSupreme last updated on 16/Feb/21

OG_1 =(a,2a)  OG_2 =(2a,a)  A_1 =A_2   OG=((ρ_1 A_1 (ai^→ +2aj^→ )+ρ_2 A_2 (2a_1 i^→ +a_2 j^→ ))/(ρ_1 A_1 +ρ_2 A_2 ))  OG=a(((ρ_1 +2ρ_2 )/(ρ_1 +ρ_2 )))i^→ +a(((2ρ_1 +ρ_2 )/(ρ_1 +ρ_2 )))j^→

$${OG}_{\mathrm{1}} =\left({a},\mathrm{2}{a}\right) \\ $$$${OG}_{\mathrm{2}} =\left(\mathrm{2}{a},{a}\right) \\ $$$${A}_{\mathrm{1}} ={A}_{\mathrm{2}} \\ $$$${OG}=\frac{\rho_{\mathrm{1}} {A}_{\mathrm{1}} \left({a}\overset{\rightarrow} {{i}}+\mathrm{2}{a}\overset{\rightarrow} {{j}}\right)+\rho_{\mathrm{2}} {A}_{\mathrm{2}} \left(\mathrm{2}{a}_{\mathrm{1}} \overset{\rightarrow} {{i}}+{a}_{\mathrm{2}} \overset{\rightarrow} {{j}}\right)}{\rho_{\mathrm{1}} {A}_{\mathrm{1}} +\rho_{\mathrm{2}} {A}_{\mathrm{2}} } \\ $$$${OG}={a}\left(\frac{\rho_{\mathrm{1}} +\mathrm{2}\rho_{\mathrm{2}} }{\rho_{\mathrm{1}} +\rho_{\mathrm{2}} }\right)\overset{\rightarrow} {{i}}+{a}\left(\frac{\mathrm{2}\rho_{\mathrm{1}} +\rho_{\mathrm{2}} }{\rho_{\mathrm{1}} +\rho_{\mathrm{2}} }\right)\overset{\rightarrow} {{j}} \\ $$

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