Question

Find the distance of the point (3, 5) from the line 2x + 3y = 14 measured parallel to the line x − 2y = 1.

Answer 1

Here,

\[\left( x_1 , y_1 \right) = A \left( 3, 5 \right)\] 

It is given that the required line is parallel to x − 2y = 1

\[\Rightarrow 2y = x - 1\]

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

\[\therefore tan\theta = \frac{1}{2} \Rightarrow sin\theta = \frac{1}{\sqrt{5}}, cos\theta = \frac{2}{\sqrt{5}}\]

So, the equation of the line is

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

\[ \Rightarrow \frac{x - 3}{\frac{2}{\sqrt{5}}} = \frac{y - 5}{\frac{1}{\sqrt{5}}}\]

\[ \Rightarrow x - 3 = 2y - 10\]

\[ \Rightarrow x - 2y + 7 = 0\]

Let line \[x - 2y + 7 = 0\] cut line 2x + 3y = 14 at P.

Let AP = r
Then, the coordinates of P are given by \[\frac{x - 3}{\frac{2}{\sqrt{5}}} = \frac{y - 5}{\frac{1}{\sqrt{5}}} = r\] \[\Rightarrow x = 3 + \frac{2r}{\sqrt{5}}, y = 5 + \frac{r}{\sqrt{5}}\]

Thus, the coordinates of P are \[\left( 3 + \frac{2r}{\sqrt{5}}, 5 + \frac{r}{\sqrt{5}} \right)\].

Clearly, P lies on the line 2x + 3y = 14.

\[\therefore 2\left( 3 + \frac{2r}{\sqrt{5}} \right) + 3\left( 5 + \frac{r}{\sqrt{5}} \right) = 14\]

\[ \Rightarrow 7 + \frac{7r}{\sqrt{5}} = 0\]

\[ \Rightarrow r = - \sqrt{5}\]

∴ AP =  \[\left| r \right|\] = \[\sqrt{5}\]