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,

$$(x_1,y_1)=A(3,5)$$ 

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 $$(3+\frac{2r}{\sqrt{5}},5+\frac{r}{\sqrt{5}})$$.

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

$$\therefore 2(3+\frac{2r}{\sqrt{5}})+3(5+\frac{r}{\sqrt{5}})=14$$

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

$$\Rightarrow r=-\sqrt{5}$$

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