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Question Number 75110 by peter frank last updated on 07/Dec/19

Answered by mind is power last updated on 08/Dec/19

$$\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$$$$\mathrm{x}=0\Rightarrow {\mathrm{y}}_0=-\frac{\mathrm{c}}{\mathrm{b}}.....\mathrm{b}\ne 0$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{y}=-\frac{\mathrm{a}}{\mathrm{b}}\mathrm{x}-\frac{\mathrm{c}}{\mathrm{a}},\mathrm{we}\mathrm{have}$$$$-\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{p}}{\cos \left(\alpha \right)}$$$$-\frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{p}}{\sin \left(\alpha \right)}$$$$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=\frac{{\mathrm{c}}^2}{{\mathrm{p}}^2}\Rightarrow \underset{-}{+}\mathrm{p}=\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\Rightarrow$$$$\mathrm{r}(\mathrm{acos}\left(\theta \right)+\mathrm{bsin}\left(\theta \right))=-\mathrm{c}$$$$\mathrm{r}=-\frac{\mathrm{c}}{\mathrm{acos}\left(\theta \right)+\mathrm{bsin}\left(\theta \right)}$$$$\cos \left(ϵ\right)=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}},\sin \left(ϵ\right)=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$$$$\Rightarrow \mathrm{r}=-\frac{\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}{\cos \left(ϵ\right)\cos \left(\theta \right)+\sin \left(ϵ\right)\sin \left(\theta \right)}$$$$\mathrm{t}=-\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\underset{-}{+}\mathrm{p}$$$$\mathrm{r}=\frac{\mathrm{p}}{\cos (\theta -ϵ)}$$$$\cos \left(ϵ\right)=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\frac{1}{{\sqrt{1+(\frac{\mathrm{b}}{\mathrm{a}}})}^2}=\frac{1}{\sqrt{1+{\mathrm{tg}}^2\left(\alpha \right)}}=\cos \left(\alpha \right)$$$$\sin \left(ϵ\right)=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\frac{1}{{\sqrt{(\frac{\mathrm{a}}{\mathrm{b}}})}^2+1}=\frac{1}{\sqrt{{\cot }^2\left(\alpha \right)+1}}=\sin \left(\alpha \right)$$$$\Rightarrow \mathrm{r}=\underset{-}{+}\frac{\mathrm{p}}{\cos (\theta -\alpha )}$$$$$$

Commented by peter frank last updated on 09/Dec/19

$$thankyouverymuch$$

Commented by peter frank last updated on 09/Dec/19

$$pleasesircanigetthe$$$$diagram$$

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