Question

Find the distance of the line 2x + y = 3 from the point (−1, −3) in the direction of the line whose slope is 1.

Answer 1

Here, 

\[\left( x_1 , y_1 \right) = A\left( - 1, - 3 \right)\] and \[tan\theta = 1 \Rightarrow sin\theta = \frac{1}{\sqrt{2}}, cos\theta = \frac{1}{\sqrt{2}}\]

So, the equation of the line is

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

\[ \Rightarrow \frac{x + 1}{\frac{1}{\sqrt{2}}} = \frac{y + 3}{\frac{1}{\sqrt{2}}}\]

\[ \Rightarrow x + 1 = y + 3\]

\[ \Rightarrow x - y - 2 = 0\]

Let line

\[x - y - 2 = 0\] cut line 2x + y = 3 at P.

Let AP = r
Then, the coordinates of P are given by \[\frac{x + 1}{cos\theta} = \frac{y + 3}{sin\theta} = r\]

\[\Rightarrow x = - 1 + rcos\theta, y = - 3 + rsin\theta\]

\[\Rightarrow x = - 1 + \frac{r}{\sqrt{2}}, y = - 3 + \frac{r}{\sqrt{2}}\]

Thus, the coordinates of P are \[\left( - 1 + \frac{r}{\sqrt{2}}, - 3 + \frac{r}{\sqrt{2}} \right)\]

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

\[\therefore 2\left( - 1 + \frac{r}{\sqrt{2}} \right) - 3 + \frac{r}{\sqrt{2}} = 3\]

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

\[ \Rightarrow \frac{3r}{\sqrt{2}} = 8\]

\[ \Rightarrow r = \frac{8\sqrt{2}}{3}\]

∴ AP = \[\frac{8\sqrt{2}}{3}\]