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: Set up the three side lines.
For A(0,3), B(-2,0), C(6,1), write the equation of each side and check the sign the opposite vertex gives. A point is inside the triangle only if it matches that same sign on all three sides.

Step 2: Side AB, tested against C.
Line AB: \(3x - 2y + 6 = 0\). At C(6,1): \(18-2+6=22>0\). For \(P(\alpha,\alpha+1)\): \(3\alpha - 2(\alpha+1)+6 = \alpha+4\), so we need \(\alpha+4>0 \Rightarrow \alpha>-4\).

Step 3: Side BC, tested against A.
Line BC: \(x - 8y + 2 = 0\). At A(0,3): \(0-24+2=-22<0\). For P: \(\alpha - 8(\alpha+1)+2 = -7\alpha-6\), so we need \(-7\alpha-6<0 \Rightarrow \alpha>-\frac{6}{7}\).

Step 4: Side CA, tested against B.
Line CA: \(x+3y-9=0\). At B(-2,0): \(-2+0-9=-11<0\). For P: \(\alpha+3(\alpha+1)-9 = 4\alpha-6\), so we need \(4\alpha-6<0 \Rightarrow \alpha<\frac{3}{2}\).

Step 5: Combine.
The tightest of \(\alpha>-4\), \(\alpha>-\frac{6}{7}\), \(\alpha<\frac{3}{2}\) gives
\[\boxed{\left(-\frac{6}{7}, \frac{3}{2}\right)}\]