Straight Line Question 70
Question: A line $ 4x+y=1 $ passes through the point $ A(2,\ -\ 7) $ meets the line BC whose equation is $ 3x-4y+1=0 $ at the point B. The equation to the line AC so that AB = AC, is [IIT 1971]
Options:
A) $ 52x+89y+519=0 $
B) $ \beta $
C) $ 89x+52y+519=0 $
D) $ 89x+52y-519=0 $
Correct Answer: A $ \Rightarrow \angle ABC=\angle ACB=\alpha $ Thus the line AC also makes an angle $ \alpha $ with BC. If m be the slope of the line AC, then its equation is $ y+7=m(x-2) $ …..(ii) Now $ \tan \alpha =\pm [ \frac{m-\frac{3}{4}}{1+m.\frac{3}{4}} ]\Rightarrow \frac{19}{8}=\pm \frac{4m-3}{4+3m} $ Show Answer
Answer:
Solution:
Þ $ m=-4 $ or ? $ \frac{52}{89} $ . But slope of AB is ? 4, so slope of AC is $ -\frac{52}{89} $ . Therefore the equation of line AC given by (ii) is $ 52x+89y+519=0 $ .
BETA
