Question

The negation of \((p∧q)\rightarrow ((p∨r)\rightarrow \,\sim q)\) is equivalent to ...

• A. \(p∧q\)
• B. \(p∧\,\sim r\)
• C. \(q∧(p∨r)\)
• D. \(\sim p∧\,\sim q\)

Hint:

Substitute the line into the pair of lines to get a quadratic and use the sum of roots.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The points A and B lie on both the line \(x + y = 3\) and the pair \(x^2 - 3xy + y^2 = 0\). Substitute \(y = 3 - x\) into the pair.

Step 2: Form the quadratic:
\[ x^2 - 3x(3 - x) + (3 - x)^2 = 0 \]
\[ x^2 - 9x + 3x^2 + 9 - 6x + x^2 = 0 \Rightarrow 5x^2 - 15x + 9 = 0 \]

Step 3: Midpoint:
The sum of the roots is \(x_1 + x_2 = 3\), so the x coordinate of the midpoint is \(\frac{3}{2}\).
Then \(y = 3 - \frac{3}{2} = \frac{3}{2}\). The midpoint is \(\left(\frac{3}{2}, \frac{3}{2}\right)\).

Final Answer:
The midpoint of AB is \(\left(\frac{3}{2}, \frac{3}{2}\right)\), option (D). \[ \boxed{\left(\frac{3}{2},\frac{3}{2}\right)} \]