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

L1 : 3x – 4y + 12 = 0L2 : 8x + 6y + 11 = 0 (α, β) lies on that angle which contain origin∴ Equation of angle bisector of that angle which contain origin is3x−4y+125=8x+6y+1110⇒2x+14y−13=0 (α, β) lies on it⇒2α+14β−13=0 …… (i)⇒3α−4β+7=0 ……. (ii)Solving (i) & (ii)α=−2325&β=5350∴α+β=750⇒100 (α+β)=14

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Straight Lines Ncert Solutions Maths class 11th Maths Class 11th

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

Option 1 -

-14

Option 2 -

Option 3 -

-22

Option 4 -

5 Views | Posted 2 months ago

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$(α, β)$ lies on that angle which contain origin

$∴$ Equation of angle bisector of that angle which contain origin is

$\frac{3 x - 4 y + 12}{5} = \frac{8 x + 6 y + 11}{10}$

$\Rightarrow 2 x + 14 y - 13 = 0$

$(α, β)$ lies on it

$\Rightarrow 2 α + 14 β - 13 = 0$ …… (i)

$\Rightarrow 3 α - 4 β + 7 = 0$ ……. (ii)

Solving (i) & (ii)

$α = - \frac{23}{25} & β = \frac{53}{50}$

$∴ α + β = \frac{7}{50}$

$\Rightarrow 100 (α + β) = 14$

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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

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