Question

Determine the value(s) of “a” for which the point $(a, a^2)$ lies inside the triangle formed by the lines: $2x+3y=1$, $x+2y=3$ and $5x-6y=1$.

• A. $(-3,-1) \cup (1/2,1)$
• B. $(-\infty,-1/3) \cup (1/2,\infty)$
• C. $(-3/2,1) \cup (1/2,1)$
• D. $(-\infty,1) \cup (1/3,6)$
• E. None of the above

Hint:

For geometry–inequality problems, always: (i) write each line in standard form, (ii) test a reference point (like the origin), (iii) compare signs for same/opposite side conditions, and (iv) intersect solution ranges.

Answer 1

The Correct Option is C

Step 1: Represent the lines.
The triangle is bounded by three lines: \[ L_1: 2x+3y-1=0, \quad L_2: x+2y-3=0, \quad L_3: 5x-6y-1=0 \] We want $(a,a^2)$ to lie inside the triangle formed by $L_1, L_2,$ and $L_3$.
Step 2: Using the origin (0,0) as a reference point.
To check whether a point lies inside the triangle, we compare the sign of each line’s equation at $(a,a^2)$ and at $(0,0)$. Depending on whether they are on the same or opposite sides of the line, we derive inequalities.
Step 3: Condition with respect to $L_1: 2x+3y-1=0$.
At $(0,0)$, $L_1(0,0) = -1$. For $(a,a^2)$: \[ L_1(a,a^2) = 2a + 3a^2 - 1 \] Since they must be opposite sides, \[ (2a+3a^2-1)(-1)<0 \quad \Rightarrow \quad 3a^2+2a-1>0 \] This gives: \[ a<-1 \quad \text{or} \quad a>\tfrac{1}{3} \quad \cdots (1) \] Step 4: Condition with respect to $L_2: x+2y-3=0$.
At $(0,0)$, $L_2(0,0)=-3$. For $(a,a^2)$: \[ L_2(a,a^2) = a+2a^2-3 \] Since they must be on the same side, \[ (a+2a^2-3)(-3)>0 \quad \Rightarrow \quad a+2a^2-3<0 \] \[ 2a^2+a-3<0 \quad \Rightarrow \quad -\tfrac{3}{2}<a<1 \quad \cdots (2) \] Step 5: Condition with respect to $L_3: 5x-6y-1=0$.
At $(0,0)$, $L_3(0,0)=-1$. For $(a,a^2)$: \[ L_3(a,a^2) = 5a - 6a^2 -1 \] Same side condition: \[ (5a-6a^2-1)(-1)>0 \quad \Rightarrow \quad 6a^2 - 5a + 1>0 \] This quadratic inequality factors to: \[ (3a-1)(2a-1)>0 \quad \Rightarrow \quad a<\tfrac{1}{3} \ \text{or} \ a>\tfrac{1}{2} \quad \cdots (3) \] Step 6: Combine inequalities.
From (1), (2), and (3): \[ a \in \left(-\tfrac{3}{2}, -1\right) \cup \left(\tfrac{1}{2},1\right) \] Final Answer: \[ \boxed{a \in (-3/2,-1) \cup (1/2,1)} \]