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, $\left(x_1,y_1\right)=A\left(2,1\right)$, $\mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

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

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-2}{{\displaystyle \cos {45}^{\circ }}}=\frac{y-1}{{\displaystyle \sin {45}^{\circ }}} \\ \Rightarrow \frac{x-2}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{y-1}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ \Rightarrow x-y-1=0 \end{gathered}$$

Let AB = r
Thus, the coordinates of B are given by

$\frac{x-2}{\cos 45°}=\frac{y-1}{\sin 45°}=r$

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

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

$$\begin{gathered} \therefore 2+\frac{r}{\sqrt{2}}+2\left(1+\frac{r}{\sqrt{2}}\right)+1=0 \\ \Rightarrow 5+\frac{3r}{\sqrt{2}}=0 \\ \Rightarrow r=-\frac{5\sqrt{2}}{3} \end{gathered}$$

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