Question Number 25767 by tawa tawa last updated on 14/Dec/17
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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)](https://www.tinkutara.com/question/Q25799.png)
$$\left(\mathrm{1}\right) \\ $$$${eqn}.\:{of}\:{BC}: \\ $$$$\frac{{y}−\mathrm{8}}{{x}+\mathrm{2}}×\frac{\mathrm{14}−\mathrm{8}}{\mathrm{2}+\mathrm{2}}=−\mathrm{1} \\ $$$$\frac{{y}−\mathrm{8}}{{x}+\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{2}}=−\mathrm{1} \\ $$$$\mathrm{3}{y}+\mathrm{2}{x}=\mathrm{20} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${C}\left({x}_{{C}} ,\mathrm{0}\right) \\ $$$$\mathrm{3}×\mathrm{0}+\mathrm{2}{x}_{{C}} =\mathrm{20} \\ $$$$\Rightarrow{x}_{{C}} =\mathrm{10} \\ $$$${C}\left(\mathrm{10},\mathrm{0}\right) \\ $$$${x}_{{D}} −{x}_{{C}} ={x}_{{A}} −{x}_{{B}} \\ $$$$\Rightarrow{x}_{{D}} =\mathrm{10}+\mathrm{2}+\mathrm{2}=\mathrm{14} \\ $$$$\Rightarrow{y}_{{D}} =\mathrm{0}+\mathrm{14}−\mathrm{8}=\mathrm{6} \\ $$$${D}\left(\mathrm{14},\mathrm{6}\right) \\ $$
Commented by tawa tawa last updated on 15/Dec/17
![I really appreciate: God bless you sir.](https://www.tinkutara.com/question/Q25801.png)
$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}:\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$