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Question

The equation of the line belonging to the family of lines (x+y)+λ(2xy+1)=0 and farthest from point (1,3) is
  1. 6x+15y+7=0
  2. 6x15y+7=0
  3. 6x+15y7=0
  4. 6x15y7=0


Solution

The correct option is B 6x15y+7=0
Given: Q=(1,3) and the lines are 
x+y=0            2xy+1=02x+x+1=0x=13y=13
So, the lines are concurrent at
P=(13,13)
The member of the family that is farthest from Q will be perpendicular to line PQ
mPQ=[3(13)][1+(13)]mPQ=10343=52

So, the slope of required line
m=1mPQ=25
Therefore, the equation of the required line is
y13=25(x+13)15y5=6x+26x15y+7=0

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