Point Inside Triangle Parameter 'a'

XAT 2010 · Quantitative Ability & Data Interpretation · Hard · Quantitative Ability

Determine the value(s) of "a" for which the point (a,a2)(a, a^2) lies inside the triangle formed by the lines: 2x+3y=12x+ 3y= 1, x+2y=3x+ 2y=3 and 5x−6y=15x-6y= 1

  1. A.

    (-3,-1) ⋃ (1/2, 1)

  2. B.

    (-∞, 1/3) ⋃ (1/2, ∞)

  3. C.

    (-3/2,-1) ⋃ (1/2, 1)

  4. D.

    (-∞, 1) ⋃ (1/3, 6)

  5. E.

    None of the above

Answer

C

Explanation

Checking the position of (a,a2)(a, a^2) relative to origin (0,0)(0,0) for all three line equations: Line 1: 2x+3y−1=02x + 3y - 1 = 0. Opposite side to origin   ⟹  (2a+3a2−1)(−1)<0  ⟹  3a2+2a−1>0  ⟹  a<−1\implies (2a + 3a^2 - 1)(-1) < 0 \implies 3a^2 + 2a - 1 > 0 \implies a < -1 or a>1/3a > 1/3. Line 2: x+2y−3=0x + 2y - 3 = 0. Same side as origin   ⟹  (a+2a2−3)(−3)>0  ⟹  2a2+a−3<0  ⟹  −3/2<a<1\implies (a + 2a^2 - 3)(-3) > 0 \implies 2a^2 + a - 3 < 0 \implies -3/2 < a < 1. Line 3: 5x−6y−1=05x - 6y - 1 = 0. Same side as origin   ⟹  (5a−6a2−1)(−1)>0  ⟹  6a2−5a+1>0  ⟹  a<1/3\implies (5a - 6a^2 - 1)(-1) > 0 \implies 6a^2 - 5a + 1 > 0 \implies a < 1/3 or a>1/2a > 1/2. Combining all three conditions yields: a∈(−3/2,−1)∪(1/2,1)a \in (-3/2, -1) \cup (1/2, 1).

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