Question

A straight line drawn through the point A (2, 1) making an angle π/4 with positive x-axis intersects another line x + 2y + 1 = 0 in the point B. Find length AB.

Answer 1

Here, 

\[\left( x_1 , y_1 \right) = A \left( 2, 1 \right)\]

\[\theta = \frac{\pi}{4}\]

So, the equation of the line passing through A (2, 1) is

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

\[ \Rightarrow \frac{x - 2}{\cos {45}^\circ} = \frac{y - 1}{\sin {45}^\circ}\]

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

\[ \Rightarrow x - y - 1 = 0\]

LetAB = r
Thus, the coordinates of B are given by

\[\frac{x - 2}{\cos45^\circ} = \frac{y - 1}{\sin45^\circ} = r\]

\[\Rightarrow x = 2 + \frac{r}{\sqrt{2}}, y = 1 + \frac{r}{\sqrt{2}}\]

Clearly, point 

\[B \left( 2 + \frac{r}{\sqrt{2}}, 1 + \frac{r}{\sqrt{2}} \right)\] lies on the line x + 2y + 1 = 0.

\[\therefore 2 + \frac{r}{\sqrt{2}} + 2\left( 1 + \frac{r}{\sqrt{2}} \right) + 1 = 0\]

\[ \Rightarrow 5 + \frac{3r}{\sqrt{2}} = 0\]

\[ \Rightarrow r = - \frac{5\sqrt{2}}{3}\] 

Hence, the length of AB is \[\frac{5\sqrt{2}}{3}\] .