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) $
Show Answer
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.