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

$\text{Here}(x_1,y_1)=A(-1,3)and$

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

$\text{The slope of the line = 1}$

$tan\theta =1$

$or\theta =\frac{\pi }{4}$

$\therefore \text{Equation of line is}$

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

$\frac{x+1}{cos\frac{\pi }{4}}=\frac{y+3}{sin\frac{\pi }{4}}=r$

$or$

$x=\frac{r}{\sqrt{2}}-1andy=\frac{r}{\sqrt{2}}-3$

$p(\frac{r}{\sqrt{2}}-1\frac{r}{\sqrt{2}}-3)\text{lie in}2x+y=3$

$\therefore 2(\frac{r}{\sqrt{2}}-1)+(\frac{r}{\sqrt{2}}-3)=3$

$\Rightarrow \frac{3r}{\sqrt{2}}=8$

$r=\frac{8\sqrt{2}}{3}$

$\text{The distance of}2x+y=3\text{from}$

$(-1,-3)is\frac{8\sqrt{2}}{3}units$