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JEE MainMathematicsStraight Lines

The area of the triangle formed by the points A(2, 1) , B(5, 5) and C(x, y) is 10 sq. units. If the point C lies on the line 2x + y = 5 , and P and Q are the two possible positions of C , then the square of the distance between P and Q is

Options

  1. A20
  2. B5
  3. C40
  4. D80

Correct answer

D. 80

Step-by-step solution

The area of the triangle with vertices A(2, 1) , B(5, 5) and C(x, y) is given by: = 1 2 | 2(5 - y) + 5(y - 1) + x(1 - 5) | = 10 | 10 - 2y + 5y - 5 - 4x | = 20 | 3y - 4x + 5 | = 20 Since C(x, y) lies on the line 2x + y = 5 , we can substitute y = 5 - 2x into the area equation: | 3(5 - 2x) - 4x + 5 | = 20 | 15 - 6x - 4x + 5 | = 20 | 20 - 10x | = 20 Dividing by 10 , we get: | 2 - x | = 2 This gives two possible cases: Case 1: 2 - x = 2 x = 0 Substituting x = 0 into y = 5 - 2x , we get y = 5 . Let this point be P(0, 5)

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