Question

The points \( (1, \frac{1}{2}) \) and \( (3, -\frac{1}{2}) \) are:

• A. In between the lines \( 2x + 3y = 6 \) and \( 2x + 3y = -6 \)
• B. On the opposite side of the line \( 2x + 3y = -6 \)
• C. On the same side of the line \( 2x + 3y = -6 \)
• D. On the same side of the line \( 2x + 3y = 6 \)

Hint:

To determine which side of the line a point lies on, substitute the point's coordinates into the equation and check the sign of the result.

Answer 1

The Correct Option is A

Step 1: Check the equation of the line. The equation of the line is \( 2x + 3y = -6 \). We need to check which side of this line the points \( (1, \frac{1}{2}) \) and \( (3, -\frac{1}{2}) \) lie on. 
Step 2: Calculate the value of \( 2x + 3y \) for each point. For the point \( (1, \frac{1}{2}) \): \[ 2(1) + 3\left(\frac{1}{2}\right) = 2 + \frac{3}{2} = \frac{7}{2}. \] For the point \( (3, -\frac{1}{2}) \): \[ 2(3) + 3\left(-\frac{1}{2}\right) = 6 - \frac{3}{2} = \frac{9}{2}. \] Both values of \( 2x + 3y \) are positive, meaning both points lie on the same side of the line. Thus, the correct answer is: \[ \boxed{1}. \]