Question:
All points inside the triangle with vertices at \((1,3)\), \((5,0)\) and \((-1,2)\) must necessarily satisfy:
All points inside the triangle with vertices at \((1,3)\), \((5,0)\) and \((-1,2)\) must necessarily satisfy:
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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.
Updated On: Jun 18, 2026
- \(3x+2y\leq 0\)
- \(3x+2y>0\)
- \(2x-3y-12>0\)
- \(2x+y-13>0\)
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The Correct Option is B
Solution and Explanation
Step 1: Check the expression \(3x+2y\) at each vertex.
The vertices of the triangle are \[ (1,3),\quad (5,0),\quad (-1,2). \] For \((1,3)\), \[ 3x+2y=3(1)+2(3)=3+6=9. \] For \((5,0)\), \[ 3x+2y=3(5)+2(0)=15. \] For \((-1,2)\), \[ 3x+2y=3(-1)+2(2)=-3+4=1. \]
Step 2: Observe the sign.
At all three vertices, \[ 3x+2y>0. \] Since the triangle is the convex region formed by these vertices, every point inside the triangle also lies on the same side of the line \[ 3x+2y=0. \] Therefore, all interior points satisfy \[ 3x+2y>0. \]
Step 3: Check why other options are not necessary.
Option (1) says \[ 3x+2y\leq 0, \] which is false because all vertices give positive values.
Option (3) says \[ 2x-3y-12>0. \] At \((5,0)\), \[ 2(5)-3(0)-12=-2<0, \] so this is not necessarily true.
Option (4) says \[ 2x+y-13>0. \] At \((5,0)\), \[ 2(5)+0-13=-3<0, \] so this is also not necessarily true.
Step 4: Final conclusion.
Hence, all points inside the triangle must satisfy \[ \boxed{3x+2y>0} \]
The vertices of the triangle are \[ (1,3),\quad (5,0),\quad (-1,2). \] For \((1,3)\), \[ 3x+2y=3(1)+2(3)=3+6=9. \] For \((5,0)\), \[ 3x+2y=3(5)+2(0)=15. \] For \((-1,2)\), \[ 3x+2y=3(-1)+2(2)=-3+4=1. \]
Step 2: Observe the sign.
At all three vertices, \[ 3x+2y>0. \] Since the triangle is the convex region formed by these vertices, every point inside the triangle also lies on the same side of the line \[ 3x+2y=0. \] Therefore, all interior points satisfy \[ 3x+2y>0. \]
Step 3: Check why other options are not necessary.
Option (1) says \[ 3x+2y\leq 0, \] which is false because all vertices give positive values.
Option (3) says \[ 2x-3y-12>0. \] At \((5,0)\), \[ 2(5)-3(0)-12=-2<0, \] so this is not necessarily true.
Option (4) says \[ 2x+y-13>0. \] At \((5,0)\), \[ 2(5)+0-13=-3<0, \] so this is also not necessarily true.
Step 4: Final conclusion.
Hence, all points inside the triangle must satisfy \[ \boxed{3x+2y>0} \]
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