Straight Line Question 71
Question: In what direction a line be drawn through the point (1, 2) so that its points of intersection with the line $ x+y=4 $ is at a distance $ \frac{\sqrt{6}}{3} $ from the given point [IIT 1966; MNR 1987]
Options:
A) $ 30^{o} $
B) $ 45^{o} $
C) $ 60^{o} $
D) $ 75^{o} $
Correct Answer: DShow Answer
Answer:
Solution:
Þ $ \frac{\sqrt{6}}{3}(\cos \theta +\sin \theta )=1 $ or $ \sin \theta +\cos \theta =\frac{3}{\sqrt{6}} $
Þ $ \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{\sqrt{3}}{2} $ , {Dividing both sides by $ \sqrt{2} $ }
Þ $ \sin (\theta +45^{o})=\sin 60^{o} $ or sin $ 120^{o} $
Þ $ \theta =15^{o} $ or $ 75^{o} $ .
BETA
