Question

A straight line passing through the point A(-1,2) has inclination of 60°and intersects the line x+y=5 at P. FindAP

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Point}\mathrm{A}\left(-1,2\right)\mathrm{inclination}60° \\ \mathrm{m}=\tan \left(60°\right)=\sqrt{3} \\ \mathrm{Eqution}\mathrm{of}\mathrm{line}\mathrm{through}\mathrm{A}\left(-1,2\right)\mathrm{and}\mathrm{slope}\mathrm{m}=\sqrt{3}\mathrm{is} \\ \mathrm{y}=\mathrm{mx}+\mathrm{c}\Rightarrow \mathrm{y}=\sqrt{3}\mathrm{x}+\mathrm{c} \\ \Rightarrow 2=-\sqrt{3}+\mathrm{c} \\ \Rightarrow \mathrm{c}=2+\sqrt{3} \\ \Rightarrow \mathrm{y}=\sqrt{3}\mathrm{x}+2+\sqrt{3} \\ \Rightarrow \sqrt{3}\mathrm{x}-\mathrm{y}=-\left(2+\sqrt{3}\right) \\ \mathrm{This}\mathrm{line}\mathrm{intersect}\mathrm{line}\mathrm{x}+\mathrm{y}=5\mathrm{at}\mathrm{P} \\ \mathrm{So},\mathrm{solving}\mathrm{these}\mathrm{two}\mathrm{equations}\mathrm{we}\mathrm{get}\mathrm{point}\mathrm{P}\left(2\sqrt{3}-3,8-2\sqrt{3}\right) \\ \mathrm{A}\left(-1,2\right)\mathrm{P}\left(2\sqrt{3}-3,8-2\sqrt{3}\right) \\ \mathrm{AP}=\sqrt{{\left(2\sqrt{3}-3+1\right)}^2+{\left(8-2\sqrt{3}-2\right)}^2} \\ =\sqrt{{\left(2\sqrt{3}-2\right)}^2+{\left(6-2\sqrt{3}\right)}^2} \\ =\sqrt{4{\left(\sqrt{3}-1\right)}^2+4{\left(3-\sqrt{3}\right)}^2} \\ =2\sqrt{{\left(\sqrt{3}-1\right)}^2+{\left(-\sqrt{3}\right)}^2{\left(\sqrt{3}-1\right)}^2} \\ =2\sqrt{{\left(\sqrt{3}-1\right)}^2\left(1+3\right)}=2\sqrt{4{\left(\sqrt{3}-1\right)}^2}=4\left(\sqrt{3}-1\right)\mathrm{units} \end{gathered}$$

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