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QJEEM22Maf73c: JEE Main 2022 Mathematics - Class 11 JEEM22Maf73c Solution: Let the point be at a unit distance from each of the two lines…

Let the point P(α,β)P(\alpha, \beta) be at a unit distance from each of the two lines L1:3x−4y+12=0L_{1}: 3 x-4 y+12=0 , and L2:8x+6y+11=0L_{2}: 8 x+6 y+11=0 . If PP lies below L1L_{1} and above L2{ }{L_{2}} , then 100(α+β)100(\alpha+\beta) is equal to :

Multiple Choice Options

  • A −- 14
  • B 42
  • C −- 22
  • D 14

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 53 English Explanation

L1:3x−4y+12=0{L_1}:3x - 4y + 12 = 0

L2:8x+6y+11=0{L_2}:8x + 6y + 11 = 0

Equation of angle bisector of L1 and L2 of angle containing origin

2(3x−4y+12)=8x+6y+112(3x - 4y + 12) = 8x + 6y + 11

2x+14y−13=02x + 14y - 13 = 0 ...... (i)

3α−4β+125=1{{3\alpha - 4\beta + 12} \over 5} = 1

⇒3α−4β+7=0\Rightarrow 3\alpha - 4\beta + 7 = 0 ...... (ii)

Solution of 2x+14y−13=02x + 14y - 13 = 0 and 3x−4y+7=03x - 4y + 7 = 0 gives the required point P(α,β), α=−2325,β=5350P(\alpha ,\beta ),\,\alpha = {{ - 23} \over {25}},\beta = {{53} \over {50}}

100(α+β)=14100(\alpha + \beta ) = 14

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