Question

If line $y-x+2=0$ is shifted parallel to itself towards the $X$-axis by a perpendicular distance of $3\sqrt{2}\mathrm{units}$, then the equation of the new line may be:

• A. $y=x+4$
• B. $y=x+1$
• C. $y=x-\left(2+3\sqrt{2}\right)$
• D. $y=x-8$

Answer 1

Answer: (A), (D)

$y=x-8$

Explanations for correct options:

Step 1: Form an equation for the new line

The equation of given line is $y-x+2=0$.

Also, it is given that the line is shifted parallel to itself towards the $X$-axis.

And, the perpendicular distance between the two lines $\left(d\right)=3\sqrt{2}\mathrm{units}$

Now, as we know, the distance $d$ between two parallel lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ is calculated by,

$d=\left|\frac{c_1-c_2}{\sqrt{{\left(a_1\right)}^2+{\left(b_1\right)}^2}}\right|$

Now, let the equation of the parallel line be $y-x+c=0$ $...\left(\mathrm{i}\right)$

Then, $3\sqrt{2}=\left|\frac{c-2}{\sqrt{{\left(-1\right)}^2+{\left(1\right)}^2}}\right|$

$\Rightarrow$ $3\sqrt{2}=\left|\frac{c-2}{\sqrt{1+1}}\right|$

$\Rightarrow$ $3\sqrt{2}=\left|\frac{c-2}{\sqrt{2}}\right|$

$\Rightarrow$ $3\sqrt{2}=\frac{\left|c-2\right|}{\sqrt{2}}$

$\Rightarrow$ $3\sqrt{2}\times \sqrt{2}=\left|c-2\right|$

$\Rightarrow$ $6=\left|c-2\right|$

Step 2: Calculate the value of $c$

From the above, we have,

$\left|c-2\right|=6$

$\Rightarrow$ $c-2=\pm 6$

$\Rightarrow$ $c-2=6$ or $c-2=-6$

$\Rightarrow$ $c=6+2$ or $c=-6+2$

$\Rightarrow$ $c=8$ or $c=-4$

Step 3: Determine the equation of the parallel line

Substituting these values of $c$ in the equation $\left(\mathrm{i}\right)$, we have

$\because y-x+c=0$

$\Rightarrow$ $y-x+8=0$ $\left[\because c=8\right]$

$\Rightarrow$ $y=x-8$

And, $y-x+\left(-4\right)=0$ $\left[\because c=-4\right]$

$\Rightarrow$ $y-x-4=0$

$\Rightarrow$ $y=x+4$

So, the required equation of the parallel line is $y=x-8$ or $y=x+4$.

Hence, options (A) and (D) are the correct options.