Question

A straight line passing through thee point A (-1,2) has inclination $\pi /3$ and intersects the line x + y = 5 at p. Find AP

Answer 1

Inclination, $\theta =\frac{\pi }{3}$

$\Rightarrow \tan \theta =\sqrt{3}$

$\Rightarrow$ Slope $=\sqrt{3}$

$\Rightarrow$ Equation $y-2=\sqrt{3}(x+1)$

$\Rightarrow y=\sqrt{3}x+\sqrt{3}+2$

$\Rightarrow p:\sqrt{3}x+\sqrt{3}+2=5-x$

$\Rightarrow x=\frac{3-\sqrt{3}}{\sqrt{3}+1}=\frac{\sqrt{3}(\sqrt{3}-1)}{\sqrt{3}+1}$

$\Rightarrow x=\frac{\sqrt{3}(\sqrt{3}-1{)}^2}{2}=\frac{\sqrt{3}(4-2\sqrt{3})}{2}=\sqrt{3}(2-\sqrt{3})$

$=2\sqrt{3}-3$

$y=\sqrt{3}\left(\sqrt{3}\right(2-\sqrt{3})+1)+2$

$=\sqrt{3}(2\sqrt{3}-2)+2=6+2-2\sqrt{3}$

$=8-2\sqrt{3}$

$\Rightarrow AP=\sqrt{(2\sqrt{3}-3+1{)}^2+(8-2\sqrt{3}-2{)}^2}$

$=\sqrt{(2\sqrt{3}-2{)}^2+(6-2\sqrt{3}{)}^2}$

$=2\sqrt{(\sqrt{3}-1{)}^2+(3-\sqrt{3}{)}^2}$

$=2\sqrt{3+1-2\sqrt{3}+9+3-6\sqrt{3}}$

$=2\sqrt{16-8\sqrt{3}}$

$=4\sqrt{4-2\sqrt{3}}$

$AP=4(\sqrt{3}-1)$.