Straight Line Question 134
The foot of the perpendicular drawn from origin on the line $ 3x+4y-5=0 $ , is [RPET 1990]
Options:
A) $ ( \frac{3}{5},2 ) $
B) $ ( \frac{3}{5},\frac{4}{5} ) $
C) $ ( -\frac{3}{5},-\frac{4}{5} ) $
D) $ ( \frac{30}{17},\frac{19}{17} ) $
Correct Answer: B $ d=\frac{5}{\sqrt{3^{2}+4^{2}}}= \frac{5}{5}=1 $ Slope of perpendicular = $ \frac{4}{3} $
Þ $ x=\pm 1.\cos \theta =\pm \frac{3}{5} $ and $ y=\pm 1.\sin \theta =\pm \frac{4}{5} $ Hence $ ( \frac{3}{5},\frac{4}{5} ) $ lies on unit circle.
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Answer:
Solution:
BETA
