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: Use symmetry:
The pair $x^2 - 3xy + y^2 = 0$ is symmetric in x and y, and so is the line $x + y = 3$. So the intersection points are mirror images across $y = x$.

Step 2: Conclude:
Therefore the midpoint lies on $y = x$ as well as on $x + y = 3$. Solving, $x = y = \frac{3}{2}$ (D). The discriminant $225 - 180 = 45 > 0$, so A and B are real.

Final Answer:
$\left(\frac{3}{2},\frac{3}{2}\right)$. \[ \boxed{\left(\frac{3}{2},\frac{3}{2}\right)} \]