Question

If the point $(\alpha, 0)$ lies inside the quadrilateral formed by lines $2x + 5y = 15, 5x - 4y = 21, 3x + 5y + 17 = 0$ and $y = x + 3$, then which of the following is true?

• A. Number of prime value(s) of $\alpha$ is 4.
• B. Number of integral value(s) of $\alpha$ is 7.
• C. Minimum integral value of $\alpha$ is -3.
• D. Maximum integral value of $\alpha$ is 5.

Answer 1

Answer: (B)

Number of integral value(s) of $\alpha$ is 7.

Answer 2

1. Find Vertices:
   Solve intersections of the lines:

   • L₁: 2x + 5y = 15 and L₄: y = x + 3
     Substitute y: $2x + 5(x + 3) = 15 \rightarrow 7x + 15 = 15 \rightarrow x = 0, y = 3$.
   • L₁ and L₂ (5x – 4y = 21):
     Solve $2x + 5y = 15$ and $5x – 4y = 21$ to get $x = 5, y = 1$.
   • L₂ and L₃ (3x + 5y + 17 = 0):
     Solve $5x – 4y = 21$ and $3x + 5y = –17$ to get $x = 1, y = –4$.
   • L₃ and L₄:
     Solve $3x + 5y = –17$ and $y = x + 3 \rightarrow 3x + 5(x + 3) = –17 \rightarrow 8x = –32 \rightarrow x = –4, y = –1$.

2. Determine α-range for (α, 0) inside the quadrilateral:
   The horizontal line y = 0 meets the quadrilateral’s boundaries on the sides:

   • Intersection with L₂: $5\alpha – 4 \cdot 0 = 21 \rightarrow \alpha = \frac{21}{5} = 4.2$
   • Intersection with L₄: $0 = \alpha + 3 \rightarrow \alpha = –3$
     Since $(\alpha, 0)$ is strictly inside,
     $-3 < \alpha < \frac{21}{5}$ (i.e., $-3 < \alpha < 4.2$).

3. Analyze Choices for α:

   • Integral values: $\alpha \in \{-2, -1, 0, 1, 2, 3, 4\} \rightarrow 7$ values.
   • Prime values: Only 2 and 3 are primes (2 values only).
   • Minimum integral value = –2 (–3 is excluded) and maximum = 4 (not 5).

Thus, only option (B) is true.