Question

Find the distance of the point (2, 3) from the line 2x − 3y + 9 = 0 measured along a line making an angle of 45° with the x-axis.

Answer 1

Here, $\left(x_1,y_1\right)=A\left(2,3\right),\mathrm{\theta }={45}^{\circ }$

So, the equation of the line passing through (2, 3) and making an angle of 45° with the x-axis is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x-2}{\cos {45}^{\circ }}=\frac{y-3}{\sin {45}^{\circ }} \\ \Rightarrow \frac{x-1}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{y-2}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ \Rightarrow x-y+1=0 \end{gathered}$$

Let x − y + 1 = 0 intersect the line 2x − 3y + 9 = 0 at point P.
Let AP = r
Then, the coordinates of P are given by

$\frac{x-2}{\cos 45°}=\frac{y-3}{\sin 45°}=r$

$\Rightarrow x=2+\frac{r}{\sqrt{2}}\mathrm{and}y=3+\frac{r}{\sqrt{2}}$

Thus, the coordinates of P are $\left(2+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}}\right)$.

Clearly, P lies on the line 2x − 3y + 9 = 0.

$$\begin{gathered} \therefore 2\left(2+\frac{r}{\sqrt{2}}\right)-3\left(3+\frac{r}{\sqrt{2}}\right)+9=0 \\ \Rightarrow 4+\frac{2r}{\sqrt{2}}-9-\frac{3r}{\sqrt{2}}+9=0 \\ \Rightarrow \frac{r}{\sqrt{2}}=4\Rightarrow r=4\sqrt{2} \end{gathered}$$

Hence, the distance of the point from the given line is $4\sqrt{2}$.