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, 

\[\left( x_1 , y_1 \right) = A \left( 2, 3 \right), \theta = {45}^\circ\]

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

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

\[ \Rightarrow \frac{x - 2}{\cos {45}^\circ} = \frac{y - 3}{\sin {45}^\circ}\]

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

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

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 \[\frac{x - 2}{\cos45^\circ} = \frac{y - 3}{\sin45^\circ} = r\]

\[\Rightarrow x = 2 + \frac{r}{\sqrt{2}}\text {  and  }y = 3 + \frac{r}{\sqrt{2}}\]

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.

\[\therefore 2\left( 2 + \frac{r}{\sqrt{2}} \right) - 3\left( 3 + \frac{r}{\sqrt{2}} \right) + 9 = 0\]

\[ \Rightarrow 4 + \frac{2r}{\sqrt{2}} - 9 - \frac{3r}{\sqrt{2}} + 9 = 0\]

\[ \Rightarrow \frac{r}{\sqrt{2}} = 4 \Rightarrow r = 4\sqrt{2}\]

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