Straight Line Question 359
Question: The equation of the line bisecting perpendicularly the segment joining the points (? 4, 6) and (8, 8) is [Karnataka CET 2003]
Options:
A) $ 6x+y-19=0 $
B) $ y=7 $
C) $ 6x+2y-19=0 $
D) $ x+2y-7=0 $
Correct Answer: A
Show Answer
Answer:
Solution:
Þ $ 6y-36=x+4 $
Þ $ 6y-x-40=0 $ ??(i) Now equation of any line perpendicular to it is $ 6x+y+\lambda =0 $ ??(ii) This line passes through the mid point of $ (-4,6) $ and $ (8,8) $ i.e., $ (2,7) $
Þ $ 6\times 2+7+\lambda =0 $
Þ $ 19+\lambda =0\Rightarrow \lambda =-19 $ From (ii) the equation of required line is $ 6x+y-19=0 $ .
BETA
