Question

The distance between the points \(P(-2, 5)\) and \(Q(5, -2)\) is

• A. \(7\sqrt{2}\)
• B. 14
• C. \(2\sqrt{7}\)
• D. 7

Hint:

Notice that the absolute horizontal difference is \(| 5 - (-2) | = 7\), and the absolute vertical difference is \(| -2 - 5 | = 7\).
Whenever the horizontal difference and vertical difference are equal (let's say they are \(k\)), the straight-line distance is always \(k\sqrt{2}\).
Since \(k = 7\) in this case, the distance is immediately \(7\sqrt{2}\).

Answer 1

The Correct Option is A

Step 1: Note the coordinates.
Point $P(x_1, y_1) = (-2, 5)$ and Point $Q(x_2, y_2) = (5, -2)$.
Step 2: Recall the Distance Formula.
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 3: Compute the differences.
$x_2 - x_1 = 5 - (-2) = 7$ and $y_2 - y_1 = -2 - 5 = -7$.
Step 4: Square and add the differences.
\[ (7)^2 + (-7)^2 = 49 + 49 = 98 \]
Step 5: Take the square root.
\[ PQ = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2} \]
Step 6: Conclusion.
The distance between points $P$ and $Q$ is $7\sqrt{2}$ units.
\[ \boxed{7\sqrt{2}} \]