JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

Let two points be A(1, –1) and B(0, 2). If a point P(x', y') be such that the area of ΔPAB = 5 sq. units and it lies on the line, 3x + y – 4λ = 0, then a value of λ is :

  1. (a) 4
  2. (b) 1
  3. (c) -3
  4. (d) 3
Answer and solution

Answer (d) 3

Solution

Maths sir · 4 steps

Write the area as a determinant in x' and y', and you get 3x'+y', exactly what the line gives you.

  1. Area formula. For P(x', y'), A(1, -1) and B(0, 2), the area of a triangle from its vertices is half the absolute value of the determinant.

    1/2| x'(-1-2) + 1(2-y') + 0(y'+1) | = 1/2| 3x' + y' - 2 |

  2. Set area to 5. 1/2|3x'+y'-2| = 5 gives |3x'+y'-2| = 10. So 3x'+y' = 12 or 3x'+y' = -8.
  3. Use the line. P lies on 3x+y-4λ=0, so 3x'+y' = 4λ. Then 4λ = 12 gives λ = 3, or 4λ = -8 gives λ = -2.
  4. So. Among the options, the possible value is λ = 3, which is option (d).

The idea to keep. When the area condition gives an expression like 3x+y and the line has the same 3x+y, substitute directly. Always keep both signs of the modulus; each sign gives a separate answer.

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 2020.