Question

Find the direction in which a straight line must be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of $\sqrt{\frac{2}{3}}$ from this point.

Answer 1

Here, $\left(x_1,y_1\right)=A\left(1,2\right)$

Let P be the point of intersection of both the lines.
∴ AP = r = $\sqrt{\frac{2}{3}}$
Let $\mathrm{\theta }$ be the slope of the line. So, the equation of the line that has slope $\mathrm{\theta }$ and passes through A (1, 2) is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-1}{\mathrm{cos\theta }}=\frac{y-2}{\mathrm{sin\theta }} \end{gathered}$$

The coordinates of P are given by

$$\begin{gathered} \frac{x-1}{\mathrm{cos\theta }}=\frac{y-2}{\mathrm{sin\theta }}=r=\sqrt{\frac{2}{3}} \\ \Rightarrow x=1+\sqrt{\frac{2}{3}}\mathrm{cos\theta },y=2+\sqrt{\frac{2}{3}}\mathrm{sin\theta } \end{gathered}$$

Thus, the coordinates of P are $\left(1+\sqrt{\frac{2}{3}}\mathrm{cos\theta },2+\sqrt{\frac{2}{3}}\mathrm{sin\theta }\right)$.

Clearly, P lies on the line x + y = 4.

$$\begin{gathered} 1+\sqrt{\frac{2}{3}}\mathrm{cos\theta }+2+\sqrt{\frac{2}{3}}\mathrm{sin\theta }=4 \\ \Rightarrow \mathrm{cos\theta }+\mathrm{sin\theta }=\frac{\sqrt{3}}{\sqrt{2}} \\ \Rightarrow {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }+2\mathrm{sin\theta cos\theta }=\frac{3}{2}\left[\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow \sin 2\mathrm{\theta }=\frac{3}{2}-1=\frac{1}{2} \\ \Rightarrow 2\mathrm{\theta }={30}^{\circ }\mathrm{or}{150}^{\circ } \\ \Rightarrow \mathrm{\theta }={15}^{\circ }\mathrm{or}{75}^{\circ } \end{gathered}$$

Hence, the direction of the lines with the positive direction of the x-axis is ${15}^{\circ }\mathrm{or}{75}^{\circ }$.