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Question Number 16542 by gourav~ last updated on 23/Jun/17

if (a/(∣Z_2 −Z_3 ∣))=(b/(∣Z_3 −Z_1 ∣))=(c/(∣Z_1 −Z_2 ∣))  Then..  find.. (a^2 /(Z_2 −Z_3 )) + (b^2 /(Z_3 −Z_1 )) + (c^2 /(Z_1 −Z_2 ))     a) 0  b) 1  c) 2  d)N.O.T

$${if}\:\frac{{a}}{\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid}=\frac{{b}}{\mid{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \mid}=\frac{{c}}{\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid}\:\:{Then}.. \\ $$$${find}..\:\frac{{a}^{\mathrm{2}} }{{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} }\:+\:\frac{{b}^{\mathrm{2}} }{{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} }\:+\:\frac{{c}^{\mathrm{2}} }{{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} }\: \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{0} \\ $$$$\left.{b}\right)\:\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{2} \\ $$$$\left.{d}\right){N}.{O}.{T} \\ $$

Commented by prakash jain last updated on 23/Jun/17

 (a/(∣Z_2 −Z_3 ∣))=(b/(∣Z_3 −Z_1 ∣))=(c/(∣Z_1 −Z_2 ∣)) =k  a=k∣Z_2 −Z_3 ∣  a^2 =k^2 (∣Z_2 −Z_3 ∣)^2 =k^2 (Z_2 −Z_3 )(Z_2 ^− −Z_3 ^− )  (a^2 /(Z_2 −Z_3 ))=k^2 (Z_2 ^− −Z_3 ^− )  ans 0.

$$\:\frac{{a}}{\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid}=\frac{{b}}{\mid{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \mid}=\frac{{c}}{\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid}\:={k} \\ $$$${a}={k}\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid \\ $$$${a}^{\mathrm{2}} ={k}^{\mathrm{2}} \left(\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid\right)^{\mathrm{2}} ={k}^{\mathrm{2}} \left({Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \right)\left(\overset{−} {{Z}}_{\mathrm{2}} −\overset{−} {{Z}}_{\mathrm{3}} \right) \\ $$$$\frac{{a}^{\mathrm{2}} }{{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} }={k}^{\mathrm{2}} \left(\overset{−} {{Z}}_{\mathrm{2}} −\overset{−} {{Z}}_{\mathrm{3}} \right) \\ $$$${ans}\:\mathrm{0}. \\ $$

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