Question

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

Answer 1

\[\text { Here, } \left( x_1 , y_1 \right) = A\left( 3, 5 \right), \tan\theta = \frac{1}{2}\]

\[ \Rightarrow sin\theta = \frac{1}{\sqrt{1^2 + 2^2}} \text { and  }cos\theta = \frac{2}{\sqrt{1^2 + 2^2}}\]

\[ \Rightarrow sin\theta = \frac{1}{\sqrt{5}}\text {  and } cos\theta = \frac{2}{\sqrt{5}}\]

So, the equation of the line passing through (3, 5) and having slope  \[\frac{1}{2}\] is

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

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

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

Let x − 2y + 7 = 0 intersect the line 2x + 3y = 14 at point P.
Let AP = r
Then, the coordinates of P are given by \[\frac{x - 3}{\frac{2}{\sqrt{5}}} = \frac{y - 5}{\frac{1}{\sqrt{5}}} = r\]

\[\Rightarrow x = 3 + \frac{2r}{\sqrt{5}} \text { and }y = 5 + \frac{r}{\sqrt{5}}\]

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.

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

\[ \Rightarrow 6 + \frac{4r}{\sqrt{5}} + 15 + \frac{3r}{\sqrt{5}} = 14\]

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

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

Hence, the distance of the point (3, 5) from the line 2x + 3y = 14 is \[\sqrt{5}\].