Question

Suppose the vertices of a triangle are given by \(A(0,3)\), \(B(-2,0)\), and \(C(6,1)\). For \((\alpha, \alpha + 1)\) to lie inside the triangle, \(\alpha\) should lie in the interval

• A. \(\left(-\frac{6}{7}, 4\right)\)
• B. \(\left(\frac{4}{5}, 4\right)\)
• C. \((-\infty, -\frac{6}{7}) \cup (4, \infty)\)
• D. \(\left(-\frac{6}{7}, \frac{3}{2}\right)\)

Hint:

To check if a point lies inside a triangle, use the area method: sum of sub-triangle areas equals area of original triangle; solve inequalities to find the interval.

Answer 1

The Correct Option is D

Step 1: Use area method.
Point \((\alpha, \alpha+1)\) lies inside triangle \(ABC\) if sum of areas of sub-triangles with this point equals area of \(ABC\).

Step 2: Area of triangle ABC.
\[ \text{Area} = \frac{1}{2} \left| 0(0-1) + (-2)(1-3) + 6(3-0) \right| = \frac{1}{2} |0 + 4 + 18| = 11 \]

Step 3: Areas of sub-triangles with P(\(\alpha,\alpha+1\)).
Use determinant formula for areas. Inequalities for being inside triangle yield \[ -\frac{6}{7} \lt \alpha \lt \frac{3}{2} \]

Step 4: Final interval.
Hence, \(\alpha\) must lie in \[ \boxed{\left(-\frac{6}{7}, \frac{3}{2}\right)} \]