UPSC NDA Math Model Paper 10

We have, 3x−4y+12=0 and 3x−4y=6 ⇒3x−4y−6=0
1=

|12−(−6)|

√32+(−4)2

= 18/5 Let, equation of line midway between the given lines 3x−4y+c=0
As, it is in the middle, distance from line 3x−4y+12=0 will be

18 5

2

=

9

5

Now, 9∕5=

|12−c|

√32+(−4)2

⇒12−c=9
⇒c=3
Hence, option (4) is correct.