JEE Main 2024 · 29 January · evening · Single correct · +4 / −1

Let A be the point of intersection of the lines 3x+2y=14,5x-y=6 and B be the point of intersection of the lines 4x+3y=8,6x+y=5. The distance of the point P(5,-2) from the line AB is

  1. (a) 13/2
  2. (b) 8
  3. (c) 5/2
  4. (d) 6
Answer and solution

Answer (d) 6

Solution

Maths sir · 4 steps

Find A and B by solving each pair of lines, write line AB, then use the distance formula.

  1. Find A. From 5x - y = 6, y = 5x - 6. Put this into 3x + 2y = 14 to get 13x - 12 = 14, so x = 2 and y = 4. So A(2, 4).
  2. Find B. From 6x + y = 5, y = 5 - 6x. Put this into 4x + 3y = 8 to get 15 - 14x = 8, so x = 1/2 and y = 2. So B(1/2, 2).
  3. Equation of AB. The slope is (4 - 2)/(2 - 1/2) = 4/3. Then y - 4 = 4/3(x - 2) gives 4x - 3y + 4 = 0. Check with B: 2 - 6 + 4 = 0.

    d = (|4(5) - 3(-2) + 4|)/√(4² + 3²) = (|20 + 6 + 4|)/5 = 30/5

  4. So. The distance of P(5, -2) from line AB is 6, which is option (d).

The idea to keep. Put the line in the form ax + by + c = 0 before using (|ax₁ + by₁ + c|)/√(a² + b²). Check the line with the second point so a sign slip does not carry through.

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 2024.