Question

The distance between the points \(P(-4, 5)\) and \(Q(-1, 2)\) is

• A. 5
• B. \(3\sqrt{2}\)
• C. 6
• D. \(2\sqrt{3}\)

Hint:

Notice that the absolute difference in \(x\) coordinates is \(| -1 - (-4) | = 3\), and the absolute difference in \(y\) coordinates is \(| 2 - 5 | = 3\).
Whenever the absolute horizontal and vertical differences are equal (let's say they are \(k\)), the straight-line distance is always \(k\sqrt{2}\).
Since \(k = 3\) here, the distance is immediately \(3\sqrt{2}\).

Answer 1

The Correct Option is B

Step 1: Identify the Coordinates.
We are given $P(-4, 5)$ and $Q(-1, 2)$. We need the distance $PQ$.
Step 2: Recall the Distance Formula.
For two points $(x_1, y_1)$ and $(x_2, y_2)$: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 3: Calculate the Differences.
\[ x_2 - x_1 = -1 - (-4) = -1 + 4 = 3 \] \[ y_2 - y_1 = 2 - 5 = -3 \]
Step 4: Apply the Formula.
\[ PQ = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \]
Step 5: Simplify the Result.
\[ \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \]
Step 6: Match with Options.
$3\sqrt{2}$ corresponds to option (2).
\[ \boxed{3\sqrt{2}} \]