Let the point P(α, β) be at a unit distance from each of the two lines L 1: 3 x -4 y +12=0, and L 2: 8 x +6 y +11=0. If P lies below L 1 and above L 2, then 100(α+β) is equal to

Question

Let the point P(α, β) be at a unit distance from each of the two lines L 1: 3 x -4 y +12=0, and L 2: 8 x +6 y +11=0. If P lies below L 1 and above L 2, then 100(α+β) is equal to (A) -14 (B) 42 (C) -22 (D) 14

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a8a5ccf0b8ad8b81748db0ad342a4232-.png" /> By observing origin and P lies in same region. L 1(0,0)>0 ; L 1(α, β)>0 ⇒ 3 α-4 β+12>0 1=|(3 α-4 β+12/5)| 3 α-4 β+12=5......(1) Similarly for L 2 L 2(0,0)>0 ; L 2(α, β)>0 1=|(8 α+6 β+11/10)| ⇒ 8 α+6 β+11=10 .......(2) Solving (1) and (2) α=-(23/25) ; β=(106/100) 100(α+β)=100((-92/100)+(106/100))=14

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

$- 14$

42

$- 22$

14

Q. Let the point P(α,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 100(α+β) is equal to

−14

42

−22

14

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

$- 14$

$- 22$