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Question: Answered & Verified by Expert
Let the point Pα,β be at a unit distance from each of the two lines L1:3x-4y+12=0, and L2:8x+6y+11=0. If P lies below L1 and above L2, then 100α+β is equal to
MathematicsStraight LinesJEE MainJEE Main 2022 (25 Jul Shift 2)
Options:
  • A -14
  • B 42
  • C -22
  • D 14
Solution:
2842 Upvotes Verified Answer
The correct answer is: 14

Given, the point Pα,β be at a unit distance from each of the two lines L1:3x-4y+12=0, and L2:8x+6y+11=0, so on plotting the diagram we get,

​​​​​​​

Now, L1:3x-4y+12=0 and L2:8x+6y+11=0

Since L1 & L2 are perpendicular so it will form square of unit length, so equation of angle bisector of L1 and L2 of angle containing origin and will pass through α,β, so

Equation of angle bisector will be, 3x-4y+1232+42=8x+6y+1182+62

23x-4y+12=8x+6y+11

2x+14y-13=0

2α+14β-13=0     i

Also given perpendicular distance is one unit so, 3α-4β+125=1

3α-4β+7=0     ii

Solving equation i & ii

 2α+14β-13=0

 3α-4β+7=0 

Pα,β,α=-2325,β=5350

So, 100α+β=14

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