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Question-25767




Question Number 25767 by tawa tawa last updated on 14/Dec/17
Answered by mrW1 last updated on 15/Dec/17
(1)  eqn. of BC:  ((y−8)/(x+2))×((14−8)/(2+2))=−1  ((y−8)/(x+2))×(3/2)=−1  3y+2x=20    (2)  C(x_C ,0)  3×0+2x_C =20  ⇒x_C =10  C(10,0)  x_D −x_C =x_A −x_B   ⇒x_D =10+2+2=14  ⇒y_D =0+14−8=6  D(14,6)
$$\left(\mathrm{1}\right) \\ $$$${eqn}.\:{of}\:{BC}: \\ $$$$\frac{{y}−\mathrm{8}}{{x}+\mathrm{2}}×\frac{\mathrm{14}−\mathrm{8}}{\mathrm{2}+\mathrm{2}}=−\mathrm{1} \\ $$$$\frac{{y}−\mathrm{8}}{{x}+\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}=−\mathrm{1} \\ $$$$\mathrm{3}{y}+\mathrm{2}{x}=\mathrm{20} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${C}\left({x}_{{C}} ,\mathrm{0}\right) \\ $$$$\mathrm{3}×\mathrm{0}+\mathrm{2}{x}_{{C}} =\mathrm{20} \\ $$$$\Rightarrow{x}_{{C}} =\mathrm{10} \\ $$$${C}\left(\mathrm{10},\mathrm{0}\right) \\ $$$${x}_{{D}} −{x}_{{C}} ={x}_{{A}} −{x}_{{B}} \\ $$$$\Rightarrow{x}_{{D}} =\mathrm{10}+\mathrm{2}+\mathrm{2}=\mathrm{14} \\ $$$$\Rightarrow{y}_{{D}} =\mathrm{0}+\mathrm{14}−\mathrm{8}=\mathrm{6} \\ $$$${D}\left(\mathrm{14},\mathrm{6}\right) \\ $$
Commented by tawa tawa last updated on 15/Dec/17
I really appreciate:  God bless you sir.
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}:\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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