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Straight Lines and Pair of Straight Lines question

2022 · 25 Jul · Shift 2 · Q30
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  5. /2022 · 25 Jul · Shift 2 · Q30

Straight Lines and Pair of Straight Lines question

2022 · 25 Jul · Shift 2 · Q30

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let the point P(α,β)P(\alpha, \beta)P(α,β) be at a unit distance from each of the two lines L1:3x−4y+12=0L_{1}: 3 x-4 y+12=0L1​:3x−4y+12=0, and L2:8x+6y+11=0L_{2}: 8 x+6 y+11=0L2​:8x+6y+11=0. If PPP lies below L1L_{1}L1​ and above L2{ }{L_{2}}L2​, then 100(α+β)100(\alpha+\beta)100(α+β) is equal to :
  1. A
    −-− 14
  2. B
    42
  3. C
    −-− 22
  4. D
    14
View written solutionFree

Correct answer: D

  1. Use the point-to-line distance formula

For a point P(α,β)P(\alpha,\beta)P(α,β) and a line ax+by+c=0ax+by+c=0ax+by+c=0, the perpendicular distance is

∣aα+bβ+c∣a2+b2.\frac{|a\alpha+b\beta+c|}{\sqrt{a^2+b^2}}.a2+b2​∣aα+bβ+c∣​.

Since PPP is at unit distance from each line, we get one equation from each line.

  1. Distance from L1:3x−4y+12=0L_1: 3x-4y+12=0L1​:3x−4y+12=0

The distance condition gives

∣3α−4β+12∣32+(−4)2=1\frac{|3\alpha-4\beta+12|}{\sqrt{3^2+(-4)^2}}=132+(−4)2​∣3α−4β+12∣​=1 ∣3α−4β+12∣5=1\frac{|3\alpha-4\beta+12|}{5}=15∣3α−4β+12∣​=1 ∣3α−4β+12∣=5.|3\alpha-4\beta+12|=5.∣3α−4β+12∣=5.

So,

3α−4β+12=±5.3\alpha-4\beta+12=\pm 5.3α−4β+12=±5.

Now use the condition: PPP lies below L1L_1L1​.

From

3x−4y+12=0  ⟹  y=34x+3,3x-4y+12=0 \implies y=\frac{3}{4}x+3,3x−4y+12=0⟹y=43​x+3,

"below the line" means

β<34α+3.\beta<\frac{3}{4}\alpha+3.β<43​α+3.

Multiplying by 444,

4β<3α+124\beta<3\alpha+124β<3α+12 3α−4β+12>0.3\alpha-4\beta+12>0.3α−4β+12>0.

Hence the quantity inside modulus is positive, so

3α−4β+12=53\alpha-4\beta+12=53α−4β+12=5 3α−4β=−7.(1)3\alpha-4\beta=-7. \qquad (1)3α−4β=−7.(1)
  1. Distance from L2:8x+6y+11=0L_2: 8x+6y+11=0L2​:8x+6y+11=0

Similarly,

∣8α+6β+11∣82+62=1\frac{|8\alpha+6\beta+11|}{\sqrt{8^2+6^2}}=182+62​∣8α+6β+11∣​=1 ∣8α+6β+11∣10=1\frac{|8\alpha+6\beta+11|}{10}=110∣8α+6β+11∣​=1 ∣8α+6β+11∣=10.|8\alpha+6\beta+11|=10.∣8α+6β+11∣=10.

So,

8α+6β+11=±10.8\alpha+6\beta+11=\pm 10.8α+6β+11=±10.

Now use the condition: PPP lies above L2L_2L2​.

From

8x+6y+11=0  ⟹  y=−43x−116,8x+6y+11=0 \implies y=-\frac{4}{3}x-\frac{11}{6},8x+6y+11=0⟹y=−34​x−611​,

"above the line" means

β>−43α−116.\beta> -\frac{4}{3}\alpha-\frac{11}{6}.β>−34​α−611​.

Multiplying by 666,

6β>−8α−116\beta>-8\alpha-116β>−8α−11 8α+6β+11>0.8\alpha+6\beta+11>0.8α+6β+11>0.

Hence,

8α+6β+11=108\alpha+6\beta+11=108α+6β+11=10 8α+6β=−1.(2)8\alpha+6\beta=-1. \qquad (2)8α+6β=−1.(2)
  1. Solve the system

We have

3α−4β=−7(1)3\alpha-4\beta=-7 \qquad (1)3α−4β=−7(1) 8α+6β=−1(2)8\alpha+6\beta=-1 \qquad (2)8α+6β=−1(2)

Multiply (1) by 333:

9α−12β=−219\alpha-12\beta=-219α−12β=−21

Multiply (2) by 222:

16α+12β=−216\alpha+12\beta=-216α+12β=−2

Add them:

25α=−2325\alpha=-2325α=−23 α=−2325.\alpha=-\frac{23}{25}.α=−2523​.

Substitute into (1):

3(−2325)−4β=−73\left(-\frac{23}{25}\right)-4\beta=-73(−2523​)−4β=−7 −6925−4β=−7-\frac{69}{25}-4\beta=-7−2569​−4β=−7 −4β=−7+6925=−10625-4\beta=-7+\frac{69}{25}=-\frac{106}{25}−4β=−7+2569​=−25106​ β=5350.\beta=\frac{53}{50}.β=5053​.
  1. Compute 100(α+β)100(\alpha+\beta)100(α+β)
α+β=−2325+5350\alpha+\beta=-\frac{23}{25}+\frac{53}{50}α+β=−2523​+5053​ α+β=−4650+5350=750.\alpha+\beta=-\frac{46}{50}+\frac{53}{50}=\frac{7}{50}.α+β=−5046​+5053​=507​.

Therefore,

100(α+β)=100⋅750=14.100(\alpha+\beta)=100\cdot \frac{7}{50}=14.100(α+β)=100⋅507​=14.
  1. Check the options
  • A: −14-14−14
  • B: 424242
  • C: −22-22−22
  • D: 141414

So the correct option is

D: 14.\boxed{\text{D: }14}.D: 14​.
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