show-that-the-area-of-the-triangle-whose-vertics-area-0-0-0-x-1-y-1-z-1-x-2-y-2-z-2-is-1-2-y-1-z-2-y-2-z-1-2-

Tinku Tara June 4, 2023 Coordinate Geometry 0 Comments

Question Number 49466 by munnabhai455111@gmail.com last updated on 07/Dec/18

show that the area of the triangle whose vertics area (0,0,0) , (x_1 ,y_1 ,z_1 ) , (x_2 ,y_(2,) z_2 ) is 1/2((√(Σ(y_1 z_2 −y_2 z_1 )^2 .))

$${show}\:{that}\:{the}\:{area}\:{of}\:{the}\:{triangle}\:{whose}\:{vertics}\:{area}\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:,\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:,\:\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2},} {z}_{\mathrm{2}} \right)\:{is}\:\mathrm{1}/\mathrm{2}\left(\sqrt{\Sigma\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right)^{\mathrm{2}} \:.}\right. \\ $$

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