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Question

If line yx+2=0 is shifted parallel to itself towards the X-axis by a perpendicular distance of 32units, then the equation of the new line may be:


A

y=x+4

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B

y=x+1

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C

y=x-2+32

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D

y=x-8

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Solution

The correct option is D

y=x-8


Explanations for correct options:

Step 1: Form an equation for the new line

The equation of given line is yx+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 d=32units

Now, as we know, the distance d between two parallel lines a1x+b1y+c1=0 and a2x+b2y+c2=0 is calculated by,

d=c1-c2a12+b12

Now, let the equation of the parallel line be y-x+c=0 ...(i)

Then, 32=c-2-12+12

32=c-21+1

32=c-22

32=c-22

32×2=c-2

6=c-2

Step 2: Calculate the value of c

From the above, we have,

c-2=6

c-2=±6

c-2=6 or c-2=-6

c=6+2 or c=-6+2

c=8 or c=-4

Step 3: Determine the equation of the parallel line

Substituting these values of c in the equation i, we have

y-x+c=0

y-x+8=0 c=8

y=x-8

And, y-x+-4=0 c=-4

y-x-4=0

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.


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