Question

A line passes through a point A (1, 2) and makes an angle of 60° with the x-axis and intersects the line x + y = 6 at the point P. Find AP.

Answer 1

Here, $\left(x_1,y_1\right)=A\left(1,2\right),\mathrm{\theta }={60}^{\circ }$

So, the equation of the line is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }}=r \\ \Rightarrow \frac{x-1}{\cos {60}^{\circ }}=\frac{y-2}{\sin {60}^{\circ }}=r \\ \Rightarrow \frac{x-1}{{\displaystyle \frac{1}{2}}}=\frac{y-2}{{\displaystyle \frac{\sqrt{3}}{2}}}=r \\ \mathrm{Here},r\mathrm{represents}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{this}\mathrm{line}\mathrm{from}\mathrm{point}A(1,2). \\ \mathrm{The}\mathrm{coordinates}\mathrm{of}\mathrm{any}\mathrm{point}P\mathrm{on}\mathrm{this}\mathrm{line}\mathrm{are}\left(1+\frac{r}{2},2+\frac{\sqrt{3}r}{2}\right). \end{gathered}$$

Clearly, P lies on the line x + y = 6
$$\begin{gathered} \therefore 1+\frac{r}{2}+2+\frac{\sqrt{3}r}{2}=6 \\ \Rightarrow \frac{\sqrt{3}r}{2}+\frac{r}{2}=3 \\ \Rightarrow r\left(\sqrt{3}+1\right)=6 \\ \Rightarrow r=\frac{6}{\sqrt{3}+1}=3\left(\sqrt{3}-1\right) \end{gathered}$$

∴ AP = $3\left(\sqrt{3}-1\right)$