JEE Main 2023 · 13 April · evening · Single correct · +4 / −1

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+2β and 2α-β are the roots of the equation :

  1. (a) x²-7x+12=0
  2. (b) x²-13x+42=0
  3. (c) x²-14x+48=0
  4. (d) x²-10x+25=0
Answer and solution

Answer (b) x²-13x+42=0

Solution

Maths sir · 4 steps

Pair up the lines to get the three vertices, average them, and then build the quadratic from sum and product.

  1. Find the vertices. Solve the lines two at a time. From 15x - y = 82, y = 15x - 82. Put this into the other two lines to get two vertices, then solve the last pair by elimination.
    Pair of linesWorkingVertex
    15x - y = 82 and 6x - 5y = -46x - 5(15x - 82) = -4 ⇒ -69x = -414(6, 8)
    15x - y = 82 and 9x + 4y = 179x + 4(15x - 82) = 17 ⇒ 69x = 345(5, -7)
    6x - 5y = -4 and 9x + 4y = 1724x - 20y = -16 plus 45x + 20y = 85 gives 69x = 69(1, 2)
  2. Centroid. Average the vertices: α = (6 + 5 + 1)/3 = 4 and β = (8 - 7 + 2)/3 = 1.
  3. The two roots. α + 2β = 4 + 2 = 6 and 2α - β = 8 - 1 = 7. Their sum is 13 and their product is 42.

    x² - (6 + 7)x + 6 × 7 = x² - 13x + 42 = 0

  4. So. The quadratic with roots 6 and 7 is x² - 13x + 42 = 0, which is option (b).

The idea to keep. Three lines make a triangle: solve them in pairs to get the vertices, and the centroid is just their average. A quadratic with roots p and q is x² - (p+q)x + pq = 0.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Straight Lines

Asked about 130 times in JEE Main since 2020. This question is from 2023.