Question

The distance of the lines $2\mathrm{x}-3\mathrm{y}=4$ from the point $\left(1,1\right)$ measured parallel to the line $\mathrm{x}+\mathrm{y}=1$ is

• A. $\sqrt{2}$
• B. $\frac{5}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $6$

Answer 1

The correct option is A

$\sqrt{2}$

Explanation for the correct option:

Finding the distance between the parallel lines:

Given, the line parallel to $\mathrm{x}+\mathrm{y}=1$and passes through $(1,1)$, is $\mathrm{x}+\mathrm{y}+\mathrm{k}=0$.

Hence, the line along the direction of $\mathrm{x}+\mathrm{y}=1$

Substituting the values of $x,y$ we get

$$\begin{gathered} 1+1+\mathrm{k}=0 \\ k=-2 \end{gathered}$$

Therefore, the line equation is line is $\mathrm{x}+\mathrm{y}-2=0$

Intersection of $\mathrm{x}+\mathrm{y}-2=0$and $2\mathrm{x}-3\mathrm{y}=4$ is $(2,0)$.

Substituting the points in equations $\left(1\right)\text{and}\left(2\right)$ we get,

$$\begin{gathered} 2\mathrm{x}+2\mathrm{y}=4\dots \dots \left(3\right) \\ 2\mathrm{x}-3\mathrm{y}=4\dots \dots \left(4\right) \\ \text{Solving the above equations we get,} \\ \mathrm{y}=0 \\ \text{Substituting}\mathrm{y}=0\text{in equation}\left(3\right) \\ 2\mathrm{x}+\left(0\right)=4 \\ x=2 \end{gathered}$$

Therefore, the distance between $\left(2,0\right)$ and $\left(1,1\right)$is,

$\begin{array}{rcl}\mathrm{d}& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-1\right)}^2+{\left(1-0\right)}^2}\\ & =& \sqrt{1^2+1^2}\\ & =& \sqrt{2}\end{array}$

Hence, option (A) is the correct answer.