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 $\mathrm{tan\theta }=1\Rightarrow \mathrm{sin\theta }=\frac{1}{\sqrt{2}},\mathrm{cos\theta }=\frac{1}{\sqrt{2}}$

So, the equation of the line is

$$\begin{gathered} \frac{x-x_1}{\mathrm{cos\theta }}=\frac{y-y_1}{\mathrm{sin\theta }} \\ \Rightarrow \frac{x+1}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{y+3}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ \Rightarrow x+1=y+3 \\ \Rightarrow x-y-2=0 \end{gathered}$$

Let line $x-y-2=0$ cut line 2x + y = 3 at P.

Let AP = r
Then, the coordinates of P are given by

$\frac{x+1}{\mathrm{cos\theta }}=\frac{y+3}{\mathrm{sin\theta }}=r$

$\Rightarrow x=-1+r\mathrm{cos\theta },y=-3+r\mathrm{sin\theta }$

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

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.

$$\begin{gathered} \therefore 2\left(-1+\frac{r}{\sqrt{2}}\right)-3+\frac{r}{\sqrt{2}}=3 \\ \Rightarrow -2-\sqrt{2}r-3+\frac{r}{\sqrt{2}}=3 \\ \Rightarrow \frac{3r}{\sqrt{2}}=8 \\ \Rightarrow r=\frac{8\sqrt{2}}{3} \end{gathered}$$

∴ AP = $\frac{8\sqrt{2}}{3}$