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NDA 1 2025 Maths, question 53

Topic: Analytical geometry

Q53

The vertices of a triangle are A(1,1)A(1,1), B(0,0)B(0,0) and C(2,0)C(2,0). The angular bisectors of the triangle meet at PP. What are the coordinates of PP?

  1. (a) (1,2−1)\left( 1,\sqrt{2} - 1 \right)Correct answer
  2. (b) (1,3−1)\left( 1,\sqrt{3} - 1 \right)
  3. (c) (1,1/2)(1,1/2)
  4. (d) (1/2,2−1)\left( 1/2,\sqrt{2} - 1 \right)

Solution.

Answer: (a) (1,2−1)\left( 1,\sqrt{2} - 1 \right)

  1. The angle bisectors meet at the incentre.
  2. Symmetry gives its xx-coordinate as 1.
  3. Its distance from the base BCBC is the inradius r=Δ/sr = \Delta/s.
  4. The area is Δ=1\Delta = 1 and the semiperimeter is s=1+2s = 1 + \sqrt{2}.
  5. Thus r=1/(1+2)=2−1r = 1/\left( 1 + \sqrt{2} \right) = \sqrt{2} - 1, giving P=(1,2−1)P = \left( 1,\sqrt{2} - 1 \right).