Question: The point on the line  $ x+y=4 $ which lie at a unit distance from the line  $ 4x+3y=10 $ , are      [IIT 1976]
Options:
A) $ (3,1),(-7,11) $
B) $ (3,1),(7,11) $
C) $ (-3,1),(-7,11) $
D) $ (1,3),(-7,11) $
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  Answer:
Correct Answer: A
Solution:
- Let the point  (h, k) then  $ h+k=4 $          ..?(i)     and  $ 1=\pm \frac{4h+3k-10}{\sqrt{4^{2}+3^{2}}}\Rightarrow 4h+3k=15 $  ?..(ii)     and  $ 4h+3k=5 $    ?..(iii)     On solving (i) and (ii); and (i) and (iii), we get the required points (3, 1) and (?7, 11).Trick: Check with options. Obviously, points (3, 1) and (-7, 11) lie on  $ x+y=4 $ and perpendicular distance of these points from  $ 4x+3y=10 $ is 1.