JEE Main 2025 · 7 April · morning · Single correct · +4 / −1

Let ABC be the triangle such that the equations of lines AB and AC be 3y-x=2 and x+y=2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC , then the area of the triangle PBC is equal to

  1. (a) 8
  2. (b) 4
  3. (c) 10
  4. (d) 6
Answer and solution

Answer (d) 6

Solution

Maths sir · 4 steps

Get A, B, C first. With BC on the x-axis, the altitude from A is vertical, so P is quick to find.

  1. Find the vertices. Put y = 0 in 3y - x = 2 to get B(-2, 0), and in x + y = 2 to get C(2, 0). Solving the two lines together gives 4y = 4, so A(1, 1).
  2. Two altitudes. BC lies on the x-axis, so the altitude from A is the vertical line x = 1. AC has slope -1, so the altitude from B has slope 1 and passes through (-2, 0): y = x + 2.
  3. Orthocentre. Put x = 1 in y = x + 2 to get P(1, 3). As a check, the altitude from C has slope -3 through (2, 0), which is y = -3(x - 2). It also passes through (1, 3).

    Area of △ PBC = 1/2 × BC × (height of P) = 1/2 × 4 × 3

  4. So. The base BC is 4 and P is 3 units above the x-axis, so the area is 6, option (d).

The idea to keep. When one side of the triangle lies on an axis, the altitude to that side is a vertical or horizontal line. Intersect it with one more altitude and you have the orthocentre.

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 2025.