Question

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

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

Answer 1

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

Equation of line parallel to $x+y=1$ is $x+y+c=0$

Let it passes through $(1,1)$ so the line is along direction of $x+y=1$

$\Rightarrow 1+1+c=0\Rightarrow c=-2\Rightarrow x+y-2=\mathrm{0.....}\left(i\right)$

Given line is $2x-3y-4=\mathrm{0....}\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow x=2,y=0$

So the point of intersection is $(2,0)$

So the distance of $(1,1)$ from $2x-3y-4=0$ along $x+y+1=0$ is

$=\sqrt{{(2-1)}^2+{(0-1)}^2}=\sqrt{2}$

So option $A$ is correct