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JEE Main 2019 · 10 Jan · Shift I

JEE Main 2019 Mathematics question on Straight Lines and Pair of Lines

If the line 3x+4y−24=03x + 4y - 24 = 0 intersects the X-axis at the point AA and the Y-axis at the point BB, then the incentre of the triangle OABOAB, where OO is the origin, is
  1. A.
    (4, 3)
  2. B.
    (3, 4)
  3. C.
    (4, 4)
  4. D.
    (2, 2)
Show answer and solution

Correct option: D

Solution

The line meets the x-axis (y=0) at A = (8, 0) and y-axis (x=0) at B = (0, 6). So triangle OAB has vertices O=(0,0), A=(8,0), B=(0,6).

Side lengths: OA = 8, OB = 6, AB = √(64+36) = 10.

Incentre = (a·xO + b·xA + c·xB)/(a+b+c) where a, b, c are sides opposite to O, A, B respectively. Side opposite O = AB = 10, side opposite A = OB = 6, side opposite B = OA = 8.

x-coordinate = (10·0 + 6·8 + 8·0)/(10+6+8) = 48/24 = 2.
y-coordinate = (10·0 + 6·0 + 8·6)/24 = 48/24 = 2.

Incentre = (2, 2).

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