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,

\[\left( x_1 , y_1 \right) = A \left( 1, 2 \right), \theta = {60}^\circ\]

So, the equation of the line is

\[\frac{x - x_1}{cos\theta} = \frac{y - y_1}{sin\theta} = r\]

\[ \Rightarrow \frac{x - 1}{\cos {60}^\circ} = \frac{y - 2}{\sin {60}^\circ} = r\]

\[ \Rightarrow \frac{x - 1}{\frac{1}{2}} = \frac{y - 2}{\frac{\sqrt{3}}{2}} = r\]

\[\text { Here, r represents the distance of any point on this line from point } A (1, 2) . \]

\[\text { The coordinates of any point P on this line are } \left( 1 + \frac{r}{2}, 2 + \frac{\sqrt{3}r}{2} \right) . \]

Clearly, P lies on the line x + y = 6

\[\therefore 1 + \frac{r}{2} + 2 + \frac{\sqrt{3}r}{2} = 6\]

\[ \Rightarrow \frac{\sqrt{3}r}{2} + \frac{r}{2} = 3\]

\[ \Rightarrow r\left( \sqrt{3} + 1 \right) = 6\]

\[ \Rightarrow r = \frac{6}{\sqrt{3} + 1} = 3\left( \sqrt{3} - 1 \right)\]

∴ AP = \[3\left( \sqrt{3} - 1 \right)\]