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JEE Main 202313 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2023 · 13 April · Evening Shift · Mathematics · Straight Lines

Let ( α, β ) be the centroid of the triangle formed by the lines...

Let (α,β)( \alpha, \beta ) be the centroid of the triangle formed by the lines 15x−y=82,6x−5y=−415x - y = 82, 6x - 5y = - 4 and 9x+4y=179x + 4y = 17. Then α+2β\alpha + 2\beta and 2α−β2\alpha - \beta are the roots of the equationLet ( α, β ) be the centroid of the triangle formed by the lines 15x - y = 82, 6x - 5y = - 4 and 9x + 4y = 17 . Then α

Options

  1. Ax2−7x+12=0{x}^{2} - 7x + 12 = 0
  2. Bx2−14x+48=0{x}^{2} - 14x + 48 = 0
  3. Cx2−13x+42=0{x}^{2} - 13x + 42 = 0
  4. Dx2−10x+25=0{x}^{2} - 10x + 25 = 0

Correct answer

C. x2−13x+42=0{x}^{2} - 13x + 42 = 0

Step-by-step solution

The given set of equations are:

15x−y=8215x - y = 82

6x−5y=−46x - 5y = - 4

9x+4y=179x + 4y = 17

Now on solving the above equation by taking two at a time and plotting the diagram we get,

Let ( α, β ) be the centroid of the triangle formed by the lines 15x - y = 82, 6x - 5y = - 4 and 9x + 4y = 17 . Then α

(α,β)≡(1+5+63,2−7+83)≡(4,1)( \alpha, \beta ) ≡ \left(\dfrac{1 + 5 + 6}{3}, \dfrac{2 - 7 + 8}{3}\right) ≡ ( 4, 1 )

Hence, α+2β=6\alpha + 2\beta = 6 and 2α−β=72\alpha - \beta = 7

The required equation is \left(x - 6\right)\left(x - 7\right) = 0</p><p>Equation</p><p>Equation{x}^{2} - 13x + 42 = 0

Hence this is the required option.

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