Question

All points inside the triangle with vertices at \((1,3)\), \((5,0)\) and \((-1,2)\) must necessarily satisfy:

• A. \(3x+2y\leq 0\)
• B. \(3x+2y>0\)
• C. \(2x-3y-12>0\)
• D. \(2x+y-13>0\)

Hint:

To check which inequality is necessarily satisfied inside a triangle, test the inequality at all vertices. If all vertices lie strictly on one side of a line, then every interior point also lies on that side.

Answer 1

The Correct Option is B

Step 1: Evaluate the expression 3x + 2y at each vertex of the triangle.
The vertices are (1, 3), (5, 0), and (-1, 2). Computing: at (1, 3) → 9; at (5, 0) → 15; at (-1, 2) → 1.

Step 2: Observe the consistent sign across all vertices.
All three vertices yield strictly positive values for 3x + 2y. Since the triangle is a convex region, every interior point must lie on the same side of the line 3x + 2y = 0 as its vertices.

Step 3: Conclude the necessary inequality.
Therefore, for any point inside the triangle, 3x + 2y>0 holds universally.

Step 4: Briefly explain why other options fail.
Option suggesting 3x + 2y ≤ 0 contradicts the positive vertex values. Options involving 2x - 3y - 12>0 and 2x + y - 13>0 fail at vertex (5, 0), where the expressions become negative.

Step 5: Final conclusion.
The correct condition is 3x + 2y>0.