Question

The distance of the point $(1,1)$ from the line $2x-3y-4=0$ measured in the direction of the line $x+y=1$ is

• A. $\sqrt{2}$ units
• B. $5\sqrt{2}$ units
• C. $\frac{1}{\sqrt{2}}$ units
• D. $2\sqrt{2}$ units

Answer 1

The correct option is A $\sqrt{2}$ units

The equation of a line through $P(1,1)$ and parallel to $x+y=1$ is
$\frac{x-1}{\cos \left(3\pi /4\right)}=\frac{y-1}{\sin \left(3\pi /4\right)}$

Let $PM=r.$ Then, the coordinates of $M$ are given by
$\frac{x-1}{\cos \left(3\pi /4\right)}=\frac{y-1}{\sin \left(3\pi /4\right)}=r$
ie., $x=1-\frac{r}{\sqrt{2}},y=1+\frac{r}{\sqrt{2}}$
$\because M$ lies on $2x-3y-4=0$
$\Rightarrow 2-\frac{2r}{\sqrt{2}}-3-\frac{3r}{\sqrt{2}}-4=0$
$\Rightarrow -5-\frac{5r}{\sqrt{2}}=0\Rightarrow r=-\sqrt{2}$
Hence, $PM=\sqrt{2}(\because r>0)$ units